b 


1.  c.  s. 

REFERENCE  LIBRARY 


A  SERIES  OF  TEXTBOOKS  PREPARED  FOR  THE  STUDENTS  OF  THE 
INTERNATIONAL  CORRESPONDENCE  SCHOOLS  AND  CONTAINING 
IN  PERMANENT  FORM  THE  INSTRUCTION  PAPERS. 
EXAMINATION  QUESTIONS,  AND  KEYS -USED 
IN  THEIR  VARIOUS  COURSES 


ARITHMETIC 
MENSURATION 
MECHANICAL  DEFINITIONS 
MECHANICAL  CALCULATIONS' 
YARN  CALCULATIONS,  COTTON 
YARN  CALCULATIONS,  WOOLEN 
AND  WORSTED 

YARN  CALCULATIONS,  GENERAL 
CLOTH  CALCULATIONS,  COTTON 
CLOTH  CALCULATIONS,  WOOLEN 
AND  WORSTED 
DRAFT  CALCULATIONS 
READING  TEXTILE  DRAWINGS 

983  E 

SCRANTON 

INTERNATIONAL  TEXTBOOK  COMPANY 


86 


Copyright,  1906,  by  International  Textbook  Company. 


Entered  at  Stationers’  Hall,  London. 


Arithmetic:  Copyright,  1901,  by  Christopher  Parkinson  Brooks.  Copyright,  1905, 
by  International  Textbook  Company.  Entered  at  Stationers’  Hall,  London. 
Mensuration:  Copyright,  1905,  by  International  Textbook  Company.  Entered 
at  Stationers’  Hall,  London. 

Mechanical  Definitions:  Copyright,  1901,  by  Christopher  Parkinson  Brooks. 
Copyright,  1905,  by  International  Textbook  Company.  Entered  at  Stationers’ 
Hall,  London. 

Mechanical  Calculations:  Copyright,  1901,  by  Christopher  Parkinson  Brooks. 
Copyright,  1905,  by  International  Textbook  Company.  Entered  at  Stationers’ 
Hall,  London. 

Yarn  Calculations,  Cotton:  Copyright,  1901,  by  Christopher  Parkinson  Brooks. 
Copyright,  1905,  by  International  Textbook  Company.  Entered  at  Stationers' 
Hall,  London. 

Yarn  Calculations,  Woolen  and  Worsted:  Copyright,  1901,  by  Christopher 
Parkinson  Brooks.  Copyright,  1905,  by  International  Textbook  Com¬ 
pany.  Entered  at  Stationers’  Hall,  London. 

Yarn  Calculations,  General:  Copyright,  1901,  by  Christopher  Parkinson  Brooks. 
Copyright,  1905,  by  International  Textbook  Company.  Entered  at  Stationers’ 
Hall,  London. 

Cloth  Calculations,  Cotton:  Copyright,  1901,  by  Christopher  Parkinson  Brooks. 
Copyright,  1905,  by  International  Textbook  Company.  Entered  at  Stationers’ 
Hall,  London. 

Cloth  Calculations,  Woolen  and  Worsted:  Copyright,  1901,  by  Christopher 
Parkinson  Brooks.  Copyright,  1905,  by  International  Textbook  Com¬ 
pany.  Entered  at  Stationers’  Hall,  London. 

Draft  Calculations:  Copyright,  1903,  by  International  Textbook  Company. 
Entered  at  Stationers’  Hall,  London. 

Reading  Textile  Drawings:  Copyright,  1905,  by  International  Textbook  Com¬ 
pany.  Entered  at  Stationers'  Hall,  London. 


All  rights  reserved. 


86 


Printed  in  the  United  States. 


16651 


THE  GETTY  RESEARCH 
INSTITUTE  LIBRARY 


PREFACE 


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entitled  to  receive  them  a  set  of  volumes  printed  and  bound 
especially  for  the  Course  for  which  the  student  enrolled. 
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cluded  to  issue  a  single  set  of  volumes,  comprising  all  our 
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Under  this  plan  some  volumes  contain  one  or  more  Papers 
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text  proper,  the  Examination  Questions  and  (for  those 
subjects  in  which  they  are  issued)  the  Answers  to  the 
Examination  Questions. 

In  preparing  these  textbooks,  it  has  been  our  constant 
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trouble^  The  utmost  pains  have  been  taken  to  avoid  and 
correct  any  and  all  ambiguous  expressions — both  those  due 
to  faulty  rhetoric  and  those  due  to  insufficiency  of  state¬ 
ment  or  explanation.  As  the  best  way  to  make  a  statement, 
explanation,  or  description  clear  is  to  give  a  picture  or  a 

iii 


IV 


PREFACE 


diagram  in  connection  with  it,  illustrations  have  been  used 
almost  without  limit.  The  illustrations  have  in  all  cases 
been  adapted  to  the  requirements  of  the  text,  and  projections 
and  sections  or  outline,  partially  shaded,  or  full-shaded 
perspectives  have  been  used,  according  to  which  will  best 
produce  the  desired  results. 

The  method  of  numbering  pages  and  articles  is  such  that 
each  part  is  complete  in  itself;  hence,  in  order  to  make  the 
indexes  intelligible,  it  was  necessary  to  give  each  part  a 
number.  This  number  is  placed  at  the  top  of  each  page,  on 
the  headline,  opposite  the  page  number;  and  to  distinguish 
it  from  the  page  number,  it  is  preceded  by  a  section 
mark  (§).  Consequently,  a  reference,  such  as  §3,  page  10, 
can  be  readily  found  by  looking  along  the  inside  edges  of 
the  headlines  until  §3  is  found,  and  then  through  §3  until 
page  10  is  found. 

International  Correspondence  Schools 


CONTENTS 


Arithmetic  Section  Page 

Definitions .  1  1 

Notation  and  Numeration  .  1  2 

Addition .  1  6 

Subtraction .  1  12 

Multiplication .  1  16 

Division  .  1  22 

Factors .  1  27 

Greatest  Common  Divisor .  1  29 

Least  Common  Multiple .  1  31 

Cancelation .  1  33 

Fractions .  2  1 

Decimals .  2  18 

Percentage .  3  1 

Interest .  3  7 

Ratio .  3  10 

Proportion  .  3  12 

Signs  of  Aggregation .  3  19 

Involution  .  4  1 

Evolution .  4  3 

Square  Root  .  4  4 

Cube  Root .  4  10 

Roots  of  Fractions .  4  15 

Table  Method  of  Extracting  Square  and 

Cube  Root .  4  16 

Denominate  Numbers .  5  1 

Measures .  5  2 

Addition  of  Compound  Quantities  ....  5  12 

Subtraction  of  Compound  Quantities  ...  5  13 

Multiplication  of  Compound  Quantities  .  .  5  14 

Division  of  Compound  Quantities  ....  5  16 


v 


VI 


CONTENTS 


Mensuration  Section 

Mensuration  of  Surfaces .  6 

Triangles .  6 

Quadrilaterals .  6 

Polygons .  6 

The  Circle  .  6 

Mensuration  of  Solids .  6 

The  Prism  .  .  .  6 

The  Cylinder .  6 

The  Pyramid  and  Cone .  6 

The  Sphere .  6 

Mensuration  of  Lumber . •  .  .  6 

Mechanical  Definitions 

Machine  Elements . 7 

Shafting .  7 

Power  Transmission .  7 

Belting .  7 

Pulleys .  7 

Rope  Transmission .  7 

Gearing .  7 

Cams .  7 

Levers .  7 

Mechanical  Calculations 

Rules  Pertaining  to  the  Transmission  of 

Power .  8 

Rules  Applying  to  Shafting .  8 

Rules  Applying  to  Speeds .  8 

Rules  Applying  to  Belts .  8 

Rope  Transmission .  8 

Rules  Applying  to  Gears .  8 

Rules  Applying  to  Levers .  8 

Yarn  Calculations,  Cotton 

Single  Yarns  . .  9 

Sizing  Roving  and  Yarn . 9 

Ply  Yarns .  9 

Beam  Calculations .  9 


CONTENTS  vii 

Yarn  Calculations,  Cotton — Continued  Section  Page 

Average  Numbers .  9  15 

Fancy  Warps .  9  16 

Twist  in  Yarns  .  9  18 

Breaking  Weight  of  Cotton  Warp  Yarn  .  .  9  19 

Metric  System  of  Numbering  Yarns  ...  9  21 

Yarn  Calculations,  Woolen  and  Worsted 

Worsted  Yarns . 10  1 

Sizing  Worsted  Yarns . 10  3 

Sizing  Worsted  Roving  . 10  6 

Ply  Yarns . 10  7 

Woolen  Yarns . .  ...  10  13 

Sizing  Woolen  Roving  and  Yarn  ....  10  15 

Ply  Yarns . 10  17 

Beam  Calculations . 10  18 

Fancy  Warps .  .10  21 

Average  Numbers . 10  23 

Metric  System  of  Numbering  Yarns  ...  10  25 

Yarn  Calculations,  General 

Single  Yarns . 11  2 

Cotton  Yarns . 11  2 

Numbering  System . 11  2 

Sizing  Cotton  Roving  and  Yarn . 11  4 

Woolen  Yarns . 11  7 

Run  System  of  Numbering . 11  7 

Cut  System  of  Numbering . 11  9 

Sizing  Woolen  Roving  and  Yarn  ....  11  9 

Worsted  Yarns . 11  11 

Sizing  Worsted  Yarns . 11  12 

Sizing  Worsted  Roving  . 11  13 

Silk  Yarns  .  11  14 

Jute,  Linen,  and  Ramie .  .11  16 

Equivalent  Counts . •  .  .  11  18 

Ply  Yarns . 11  20 

Woolen  and  Worsted  Ply  Yarns . 11  26 

Ply  Yarns  of  Different  Materials  ....  11  27 

Diameter  of  Yarns  . . 11  29 


Vlll 


CONTENTS 


Yarn  Calculations,  General — Continued  Section  Page 

Beamed  Yarns . 11  30 

Average  Numbers  . 11  33 

Fancy  Warps . 11  35 

Metric  System  of  Numbering  Yarns  .  .  11  38 

Cloth  Calculations,  Cotton 

Calculations  Necessary  for  Cloth  Produc¬ 
tion  . 12  1 

Harness  Calculations . 12  3 

Reeds  . 12  4 

Calculations  for  Warp  Yarns . 12  5 

Calculations  for  Filling  Yarn . 12  10 

Finding  Yards  per  Pound . 12  13 

Figuring  Particulars  From  Cloth  Samples  .  12  16 

Finding  Hanks  of  Warp  Yarn . 12  18 

Finding  Hanks  of  Filling . 12  20 

Counts  of  Filling  to  Preserve  Yards  per 

Pound . 12  22 

Average  Sley . 12  23 

Short  Rules  Used  in  Obtaining  Particu¬ 
lars  of  Cloth  Samples  .  ■ . 12  24 

Cloth  Calculations,  Woolen  and  Worsted 

Calculations  Necessary  for  Cloth  Produc¬ 
tion  . 13  1 

Construction  of  Fabrics . 13  2 

Harness  Calculations . 13  4 

Reed  Calculations . 13  5 

Contraction  in  Weaving  . 13  9 

Shrinkage  in  Finishing . 13  11 

Fancy  Patterns .  ....  13  14 

Figuring  Particulars  From  Cloth  Samples  13  21 

Draft  Calculations 

Drafting  . 15  1 

Drafting  With  Common  Rolls . 15  5 

Gearing  of  Rolls . 15  9 

Driving  and  Driven  Gears  .  • . 15  11 


CONTENTS 


IX 


Draft  Calculations — Continued  Section 

Calculating  Draft  for  Common  Rolls  .  .  15 

Break  Draft . 15 

Metallic  Rolls . 15 

Drafting  With  Special  Reference  to  the  Mill  15 

Reading  Textile  Drawings 

Representation  of  Objects . 91 

Kinds  of  Drawings . 91 

Mechanical  Drawings . 91 

Elevations  or  Side  Views . 91 

Plans . 91 

Sectional  Views . 91 

Detail  and  Assembly  Drawings . 91 

Perspective  Views  .  .  .  • . 91 

Diagrammatic  Views . 91 

Lines  Used  on  Drawings . 91 

Conventional  Methods  of  Representing 
Objects . 91 


Page 

12 

17 

20 

25 


1 

2 

3 

6 

7 

11 

17 

18 
20 
22 

28 


ARITHMETIC 

(PART  1) 


DEFINITIONS 

1.  Arithmetic  may  be  defined  as  the  science  of  numbers 
and  the  art  of  computing  with  them. 

2.  A  unit  is  a  single  thing,  or  one,  as  one  mill,  one  loom, 
or  it  may  be  used  collectively,  as  one  dozen,  which  really 
means  twelve  articles,  or  one  hank,  which  is  a  definite  number 
of  yards. 

3.  A-  number  is  a  unit  or  a  collection  of  units,  as  one, 
three,  five,  seven. 

4.  A  concrete  number  is  a  number  that  refers  to  some 
particular  article  or  quantity,  as  four  spindles,  seven  boilers, 
three  gallons. 

5.  An  abstract  number  is  a  number  that  has  no  refer¬ 
ence  to  any  particular  article  or  quantity,  as  three,  five,  seven, 
nine. 

6.  The  unit  of  a  number  is  one  of  the  same  kind  as 
the  collection  of  units  represented  by  the  number.  The  unit 
of  twelve  is  one ,  of  fifty  looms  is  one  loom,  of  ten  dollars  is 
one  dollar. 

7.  TAke  numbers  are  numbers  that  refer  to  units  of  the 
same  kind,  as  three  yards  and  six  yards,  five  quarts  and  eight 
quarts. 

8.  Unlike  numbers  are  numbers  that  refer  to  units  of 
different  kinds,  as  two  miles  and  ten  minutes ,  four  spools  and 
six  shuttles. 

For  notice  of  copyright .  see  page  immediately  following  the  title  page 

?r 


2 


ARITHMETIC 


§1 


NOTATION  AND  NUMERATION 

9.  Numbers  are  expressed  in  three  ways:  (1)  by  words; 
(2)  by  figures;  (3)  by  letters. 

10.  Notation  is  the  art  of  expressing  numbers  by 
figures  or  letters. 

11.  Numeration  is  the  art  of  expressing  in  words  num¬ 
bers  that  have  been  previously  expressed  by  figures  or  letters. 

12.  Arabic  notation,  which  is  commonly  used  for 
arithmetical  calculations,  employs  ten  characters ,  or  signs , 
called  figures  for  expressing  numbers.  These,  with  their 
names,  are  as  follows: 

Figures  1234567  89  0 

Names  one  two  three  four  five  six  seven  eight  nine  naught , 

cipher , 
or  zero 

The  last  character  (0)  is  called  naught,  cipher,  or  zero, 
and  when  standing  alone  has  no  value. 

The  other  nine  figures  are  called  digits,  and  each  has  an 
independent  value  of  its  own.  The  cipher  is  not  a  digit; 
hence,  whenever  the  word  digit  is  used,  it  refers  to  one  of 
the  first  nine  figures  above  named. 

Any  whole  number  is  called  an  integer,  or  integral 
number. 

13.  As  there  are  only  ten  figures,  it  is  evident  that  in 
expressing  numbers  each  figure  may  have  a  different  value 
at  different  times.  In  other  words,  figures  have  simple  values 
and  local ,  or  relative ,  values,  according  to  their  position  in 
relation  to  other  figures. 

14.  The  simple  value  of  a  figure  is  its  value  as  a  digit 
when  standing  alone  or,  in  a  collection  of  figures,  when  stand¬ 
ing  at  the  right  of  the  other  figures. 


§1 


ARITHMETIC 


3 


15.  The  local,  or  relative,  value  of  a  figure  is  the  value 
it  expresses  when  placed  at  the  left  of  other  figures. 

16.  The  particular  position  that  a  figure  occupies  with 
relation  to  other  figures  is  called  its  place;  thus,  in  45 
(forty-five)  the  5  occupies  the  first  place  counting  from  the 
right,  and  the  4. the  second  place. 

17.  The  local  size,  or  value,  of  the  units  of  a  figure  differs 
according  to  the  place  they  occupy;  thus,  in  the  number  566 
[five  hundred  (and)  sixty-six],  each  of  the  figures  without 
regard  to  its  place  represents  units;  but  the  6  occupying  the 
first  place,  counting  from  right  to  left,  represents  only 
6  single  units,  while  the  6  occupying  the  second  place  repre¬ 
sents  6  tens,  or  6  units  each  ten  times  the  size,  or  value,  of 
a  unit  of  the  first  place,  and  the  5  occupying  the  third  place 
from  the  right  represents  5  hundreds,  or  5  units  each  one 
hundred  times  the  size,  or  value,  of  a  unit  of  the  first  place, 
and  ten  times  the  value  of  a  unit  of  the  second  place. 

An  illustration  of  the  increasing  value  of  figures  due  to  their 
local,  or  relative,  place  in  relation  to  the  unit  place  is  as  follows: 

The  figure  7  standing  alone  represents  seven 
units;  thus . . . .  v  7 

If  another  7  is  placed  at  the  left  of  the  previous 
figure,  its  value  will  not  be  seven  units,  but  seven 
tens;  i.  e.,  ten  times  seven  units,  or  seventy  units, 
and  the  whole  number  will  be  the  sum  of  seventy 
and  seven,  or  seventy-seven;  thus . .  .  77 

If  another  7  is  placed  at  the  left  of  the  previous 
one,  its  value  will  be  seven  hundreds;  i.  e.,  one 
hundred  times  the  value  of  the  seven  in  the  first 
place  or  ten  times  the  value  of  the  seven  in  the 
second  place,  and  the  number  will  be  the  sum  of 
seven  hundred,  seventy,  and  seven,  or  seven 
hundred  (and)  seventy-seven;  thus .  777 

If  still  another  7  is  added  its  value  will  be  seven 
thojisands;  i.  e.,  ten  times  the  seven  in  the  third 
place,  one  hundred  times  the  seven  in  the  second 
place,  and  one  thousand  times  the  value  of  the 


4 


ARITHMETIC 


§1 


seven  in  the  first,  or  units,  place,  and  the  number 
will  be  the  sum  of  seven  thousand,  seven  hundred, 
seventy,  and  seven,  or  seven  thousand  seven 
hundred  (and)  seventy-seven;  thus .  7777 

Another  7  will  increase  the  number  seven  tens  of 
thousands ,  and  it  will  read  seventy-seven  thousand 
seven  hundred  (and)  seventy-seven;  thus  ....  77777 

Another  7  will  increase  the  number  seven 
hundreds  of  thousands ,  and  it  will  read  seven  hun¬ 
dred  (and)  seventy-seven  thousand  seven  hundred 
(and)  seventy-seven;  thus  ....  .......  777777 

Another  7  added  in  the  seventh  place,  beginning 
with  the  unit  and  counting  to  the  left,  will  increase 
the  number  seven  millions ,  and  it  will  then  read 
seven  million  seven  hundred  (and)  seventy-seven- 
thousand  seven  hundred  (and)  seventy-seven;  thus  7777777 

18.  In  general,  the  following  law  may  be  drawn  from 
the  above: 

Law. — The  value  expressed  by  any  figure  is  always  increased 
tenfold  each  time  it  is  removed  one  place  to  the  left. 

19.  The  cipher  (0)  has  no  value  in  itself  but  when  used 
in  conjunction  with  other  figures  it  is  useful  in  determining  the 
place  of  such  other  figures.  To  represent  the  number  six  hun¬ 
dred  seven  only  two  digits  (see  Art.  12)  are  necessary,  one  to 
represent  six  hundred  units  by  being  placed  in  the  third ,  or 
hundreds ,  place,  and  one  to  represent  the  seven  units.  The  tens , 
or  second ,  place  is  represented  by  a  cipher ;  thus,  607.  If  the 
two  digits  were  placed  together,  thus  67,  the  6  would  be  in 
the  tens ,  or  second,  place  and  the  number  would  be  sixty-seven. 

If  there  were  tivo  ciphers  between  the  6  and  7,  the  6  would 
be  thrown  into  the  thousands  place  and  the  number  would  be 
composed  of  thousands  and  units ,  reading  six  thousand  seven , 
and  written  6,007.  If  the  number  were  six  hundred  seventy , 
it  would  have  the  cipher  at  the  right-hand  side  to  show  that 
there  were  no  units.  In  this  case,  the  number  would  be 
composed  of  hundreds  and  tens ,  and  would  be  read  six  hun¬ 
dred  seventy ,  which  is  written  670. 


§1 


ARITHMETIC 


5 


20.  In  writing  numbers  it  is  customary  to  separate 
them  by  means  of  commas  into  periods  of  three  figures 
each,  beginning  at  the  right.  These  groups  of  three  figures 
are  called,  respectively,  the  periods  of  units,  of  thousands ,  of 
millions ,  of  billions ,  etc.  .  • 

A  clear  understanding  of  the  method  of  separating  a 
number  into  periods,  the  names  of  the  periods,  and  the 
names  of  the  respective  places  of  each  period  may  be 
obtained  from  the  following  example,  showing  how  the 
number  466,735,438,894,278  is  divided  into  periods. 


Period 

of 

Period  of 

Period 

of 

Period  of 

Period  of 

Trillions 

Billions 

Millions 

Thousands 

Units 

in 

XJ 

G 

in 

g 

o 

C/3 

a 

C/3 

G 

o 

Gj 

in 

G 

in 

•  H 

o 

•  H 

o 

TJ 

\ - < 

C/3 

r— H 

r-H 

in 

a 

o 

.G 

G 

u 

EH 

c 

_o 

•  rH 

m 

C/3 

a 

o 

•  r- t 

§ 

Eh 

G 

in 

M— 1 

M— 1 

•  rH 

•  rH 

F  < 

M— 1 

o 

o 

•  rH 

u 

o 

•  t-H 

o 

•  rH 

o 

.G 

in 

C/3 

EH 

C/3 

PQ 

n 

in 

Eh 

in 

C/3 

H3 

TD 

e 

rO 

<D 

U 

M-H 

O 

a 

o 

0) 

J-H 

M-l 

O 

in 

G 

<D 

M— 1 

o 

a 

<D 

O 

G 

C/3 

CD 

g 

C/3 

g 

•  rH 

T3 

G 

C/3 

G 

o 

■H 

a 

in 

G 

o 

•  rH 

'O 

C 

in 

G 

G 

O 

T3 

G 

C/3 

G 

in 

•  r—t 

CD 

u 

G 

a) 

<u 

•  rH 

G 

CD 

.G 

G 

CD 

rH 

H 

Eh 

K 

Eh 

PQ 

Eh 

S 

K 

Eh 

Eh 

EC 

Eh 

4 

6 

6, 

7 

3 

5, 

4 

3 

8, 

8 

9 

4, 

2 

7 

8 

The  value  of  the  number  expressed  in  the  above  table  is 
four  hundred  (and)  sixty-six  trillions  seven  hundred  (and) 
thirty-five  billio?is  four  hundred  (and)  thirty-eight  millions 
eight  hundred  (and)  ninety-four  thousands  two  hundred  (and) 
seventy-eight.  When  a  line  of  figures  is  read  in  this  manner 
it  is  called  numeration,  and  if  the  numeration  be  changed 
back  to  figures,  it  is  called  notation.  For  instance, 
expressing  the  following  number  in  figures, 

68,947,264, 

is  its  notation,  while  its  numeration  is  sixty-eight  millions 
nine  hundred  (and)  forty-seven  thousands  two  hundred 
(and)  sixty-four. 


6 


ARITHMETIC 


§1 


21.  It  is  customary  both  when  reading  and  writing  num¬ 
bers  to  leave  out  the  and  that  is  placed  in  parentheses  in  the 
preceding  paragraphs  and  also  to  leave  the  s  off  the  words 
millions,  thousands,  etc.;  hence,  the  above  would  be 
expressed  sixty-eight  million  nine  hundred  forty-seven 
thousand  two  hundred  sixty-four. 

The  periods  above  trillions  in  their  order  are  quadrillions, 
quintillions,  sextillions,  septillions,  octillions,  nonillions, 
decillions,  etc. 

22.  The  four  fundamental  processes  of  arithmetic  are 
addition ,  subtraction ,  multiplication,  and  division.  They  are 
called  fundamental  processes  because  all  other  arithmetical 
computations  and  processes  are  based  on  them. 


ADDITION 

23.  Addition  may  be  defined  as  the  process  of  finding  a 
number  (called  the  sum)  that  expresses  th o.  aggregate  number 
of  units  in  two  or  more  smaller  numbers. 

Only  like  numbers  can  be  added;  unlike  numbers  cannot 
be  added.  For  instance,  three  maybe  added  to  seven;  three 
looms  may  be  added  to  seven  looms;  but  it  is  clearly 
impossible  to  add  two  dollars  and  three  feet,  or  three 
wrenches  and  five  quarts. 

24.  The  sign  of  addition  is  an  erect  cross  written 
thus  +,  and  is  read  plus,  meaning  added  to.  Thus,  the 
expression  9  added  to  7  would  be  written  9  +  7,  and  would 
be  read  nine  plus  seven. 

25.  The  sign  of  equality  is  represented  by  two  parallel 
horizontal  lines  thus  =,  and  is  read  equals  or  is  equal  to, 
signifying  that  the  numerical  expression  before  the  sign 
is  equal  to  the  expression  following  the  sign;  thus,  9  +  7 
=  16  is  read  nine  plus  seven  equals  sixteen,  the  latter 
number  being  the  sum  of  the  two  previous  numbers.  The 
sum  is  also  known  as  the  total. 


§1 


ARITHMETIC 


7 


26.  The  following  table  gives  the  sum  of  any  two  num¬ 
bers  from  1  to  12: 


I 

and 

i  is 

2 

2  and 

i  is 

3 

3  and 

i  is 

4 

4  and 

1  is 

5 

I 

and 

2  is 

3 

2  and 

2  is 

4 

3  and 

2  is 

5 

4  and 

2  is 

6 

I 

and 

3  is 

4 

2  and 

3  is 

5 

3  and 

3  is 

6 

4  and 

3  is 

7 

I 

and 

4  is 

5 

2  and 

4  is 

6 

3  and 

4  is 

7 

4  and 

4  is 

8 

I 

and 

5  is 

6 

2  and 

5  is 

7 

3  and 

5  is 

8 

4  and 

5  is 

9 

I 

and 

6  is 

7 

2  and 

6  is 

8 

3  and 

6  is 

9 

4  and 

6  is 

10 

I 

and 

7  is 

8 

2  and 

7  is 

9 

3  and 

7  is 

10 

4  and 

7  is 

11 

I 

and 

8  is 

9 

2  and 

8  is 

IO 

3  and 

8  is 

11 

4  and 

8  is 

12 

I 

and 

9  is 

IO 

2  and 

9  is 

ii 

3  and 

9  is 

12 

4  and 

9  is 

13 

I 

and 

io  is 

ii 

2  and 

io  is 

12 

3  and 

io  is 

13 

4  and 

10  is 

14 

I 

and 

ii  is 

12 

2  and 

ii  is 

13 

3  and 

ii  is 

14 

4  and 

11  is 

15 

I 

and 

12  is 

13 

2  and 

12  is 

14 

3  and 

12  is 

15 

4  and 

12  is  16 

5 

and 

i  is 

6 

6  and 

i  is 

7 

7  and 

i  is 

8 

8  and 

1  is 

9 

5 

and 

2  is 

7 

6  and 

2  is 

8 

7  and 

2  is 

9 

8  and 

2  is 

10 

5 

and 

3  is 

8 

6  and 

3  is 

9 

7  and 

3  is 

10 

8  and 

3  is 

11 

5 

and 

4  is 

9 

6  and 

4  is 

IO 

7  and 

4  is 

1 1 

8  and 

4  is 

12 

5 

and 

5  is 

IO 

6  and 

5  is 

ii 

7  and 

5  is 

12 

8  and 

5  is 

13 

5 

and 

6  is 

ii 

6  and 

6  is 

12 

7  and 

6  is 

13 

8  and 

6  is 

14 

5 

and 

7  is 

12 

6  and 

7  is 

13 

7  and 

7  is 

14 

8  and 

7  is 

15 

5 

and 

8  is 

13 

6  and 

8  is 

14 

7  and 

8  is 

15 

8  and 

8  is 

16 

5 

and 

9  is 

14 

6  and 

9  is 

15 

7  and 

9  is 

16 

8  and 

9  is 

17 

5 

and 

io  is 

15 

6  and 

io  is 

16 

7  and 

io  is 

17 

8  and 

10  is 

18 

5 

and 

ii  is 

16 

6  and 

ii  is 

17 

7  and 

ii  is 

18 

8  and 

1 1  is 

19 

5 

and 

12  is 

17 

6  and 

12  is 

18 

7  and 

12  is 

19 

8  and 

12  is 

20 

9 

and 

i  is 

IO 

io  and 

i  is 

1 1 

ii  and 

i  is 

12 

12  and 

1  is 

13 

9 

and 

2  is 

1 1 

io  and 

2  is 

12 

ii  and 

2  is 

13 

12  and 

2  is 

M 

9 

and 

3  is 

12 

io  and 

3  is 

13 

ii  and 

3  is 

14 

12  and 

3  is 

15 

9 

and 

4  is 

13 

io  and 

4  is 

14 

ii  and 

4  is 

15 

12  and 

4  is 

16 

9 

and 

5  is 

14 

io  and 

5  is 

15 

ii  and 

5  is 

16 

12  and 

5  is 

17 

9 

and 

6  is 

15 

io  and 

6  is 

16 

ii  and 

6  is 

17 

12  and 

6  is 

18 

9 

and 

7  is 

16 

io  and 

7  is 

17 

ii  and 

7  is 

18 

12  and 

7  is 

19 

9 

and 

8  is 

17 

io  and 

8  is 

18 

ii  and 

8  is  jg 

12  and 

8  is 

20 

9 

and 

9  is 

18 

io  and 

9  is 

19 

ii  and 

9  is 

20 

12  and 

9  is 

21 

9 

and 

io  is 

19 

io  and 

io  is 

20 

ii  and 

io  is 

21 

12  and 

10  is 

22 

9 

and 

ii  is 

20 

io  and 

ii  is 

21 

ii  and 

ii  is 

22 

12  and 

11  is 

23 

9 

and 

12  is 

21 

io  and 

12  is 

22 

ii  and 

12  is  23 

12  and 

12  is  24 

This  table  should  be  carefully  committed  to  memory. 
Since  0  has  no  value,  the  sum  of  any  number  and  0  is  the 
number  itself;  thus  17  and  0  is  17. 

27.  When  two  or  more  numbers  are  to  be  added,  they 
are  placed  under  each  other  in  such  a  position  that  all  the 
figures  of  the  same  order  shall  be  in  the  same  column;  or, 
in  other  words,  units  must  be  placed  under  units,  tens  under 


8 


ARITHMETIC 


§1 


tens,  hundreds  under  hundreds ,  and  so  on.  Beneath  the  last, 
or  bottom,  number  a  line  is  drawn  and  the  result  of  the  addi¬ 
tion,  or  the  sum  of  the  numbers,  is  placed  beneath  this  line. 

Example  1. — What  is  the  sum  of  131,  222,  21,  2,  and  413? 

Solution.—  131 

222 

21 

2 

413 

sum  7  8  9  Ans. 

Explanation. — After  placing  the  numbers  in  proper  order, 
begin  at  the  bottom  of  the  right-hand,  or  units,  column,  and 
add,  mentally  repeating  the  different  sums.  Thus,  three  and 
two  is  five  and  one  is  six  and  two  is  eight  and  one  is  nine, 
the  sum  of  the  numbers  in  units  column.  Place  the  9 
directly  beneath  as  the  first,  or  units,  figure  in  the  sum. 

The  sum  of  the  numbers  in  the  next,  or  tens,  column 
equals  8  tens,  which  is  the  second,  or  tens,  figure  in  the  sum. 

The  sum  of  the  numbers  in  the  next,  or  hundreds,  column 
equals  7  hundreds,  which  is  the  third,  or  hundreds,  figure  in 
the  sum. 

The  sum,  or  answer,  is  789. 

Example  2. — What  is  the  sum  of  425,  36,  9,215,  4,  and  907? 


(a)  (b) 

Solution  . —  4  2  5  4  2  5 

36  36 

9215  9215 

4  4 

907  907 

27  27 

60  6 

1500  15 

9000  9 


sum  105  87  Ans.  10587  Ans. 


Explanation. — The  sum  of  the  numbers  in  the  first,  or 
units,  column  is  seven  and  four  is  eleven  and  five  is  sixteen 
and  six  is  twenty-two  and  five  is  twenty-seven,  or  27  units; 
i.  e.,  two  tens  and  seven  units.  Write  27,  as  shown.  The 


§1 


ARITHMETIC 


9 


sum  of  the  numbers  in  the  second,  or  tens,  column  is  six 
tens,  or  60.  Write  60  underneath  27,  as  shown  in  (a).  The 
sum  of  the  numbers  in  the  third,  or  hundreds,  column  is 
15  hundreds,  or  1,500.  Write  1,500  under  the  two  preceding- 
results,  as  shown.  There  is  only  one  number  in  the  fourth, 
*or  thousands,  column,  9,  which  represents  9,000.  Write 
9,000  under  the  three  preceding  results.  Adding  these  four 
results,  the  sum  is  10,587,  which  is  the  sum  of  425,  36,  9,215, 
4,  and  907.  In  practice,  the  ciphers  on  the  right  of  the 
several  sums  are  omitted,  as  in  (/>). 

Note. — It  frequently  happens,  when  adding  a  long  column  of  figures, 
that  the  sum  of  two  numbers,  one  of  which  does  not  occur  in  the 
addition  table,  is  required.  Thus,  in  the  first  column  of  example  2  ( a ) 
the  sum  of  16  and  6  was  required.  We  know  from  the  table  that 
6  +  6  =  12;  hence,  the  first  figure  of  the  sum  is  2.  Now,  the  sum  of  any 
number  less  than  20  and  of  any  number  less  than  10  must  be  less  than  30, 
since  20  +  10  =  30;  therefore,  the  sum  is  22.  Consequently,  in  cases  of 
this  kind,  add  the  first  figure  of  the  larger  number  to  the  smaller  number, 
and  if  the  result  is  greater  than  9,  increase  the  second  figure  of  the 
larger  number  by  1.  Thus,  44  +  7  —  ?  4  +  7  =  11;  hence,  44  +  7  =  51. 

The  addition  may  also  be  performed  as  follows: 

4  2  5 
3  6 
9  2  15 
4 

9  0  7 

sum  1  0  5  8  7  Ans. 

Explanation. — The  sum  of  the  numbers  in  the  units 
column  is  27  units,  or  2  tens  and  7  units.  Write  the  7  units  as 
the  first,  or  right-hand  figure,  in  the  sum.  Reserve  the  2  tens 
and  add  them  to  the  figures  in  the  tens  column.  The  sum  of 
the  figures  in  the  tens  column,  plus  the  2  tens  reserved  and 
carried  from  the  units  column  is  8,  which  is  written  down  as 
the  second  figure  in  the  sum.  There  is  nothing  to  carry  to 
the  next  column,  because  8  is  less  than  10.  The  sum  of  the 
numbers  in  the  next  column  is  15  hundreds,  or  1  thousand 
and  5  hundreds.  Write  down  the  5  as  the  third,  or  hundreds, 
figure  in  the  sum  and  carry  the  1  to  the  next  column.  1  +  9 
=  10,  which  is  written  down  at  the  left  of  the  other  figures. 


10 


ARITHMETIC 


1 


This  method  saves  space  and  figures,  but  that  shown  in 
example  2  is  to  be  preferred  when  adding  a  long  column. 

Example  3. — Add  the  numbers  in  the  column  below: 

.Solution  .  —  8  9  0 

82 

90 

393 

281 

80 

770 

83 
492 

80 

383 

84 
191 

sunt  3  8  9  9  Ans. 

Explanation. — The  sum  of  the  digits  in  the  first  column 
equals  19  units,  or  1  ten  and  9  units.  Write  down  the  9  and 
carry  1  to  the  next  column.  The  sum  of  the  digits  in  the 
second  column  +  1  =  109  tens,  or  10  hundreds  and  9  tens. 
Write  down  the  9  and  carry  the  10  to  the  next  column. 
The  sum  of  the  digits  in  this  column  plus  the  10  carried  is  38. 
The  entire  sum  is  3,899. 

28.  The  process  of  mentally  adding  the  second  figure  of 
the  sum  of  a  column  of  figures  in  with  the  next  column  is 

'  i  .  .  r 

called  carrying. 

In  some  cases,  where  long  columns  of  figures  are  being 
added,  it  is  necessary  to  add  tzvo  or  more  figures  in  with 
the  next  column.  To  illustrate  this  point,  suppose  that  in 
a  problem  in  addition  the  sum  of  the  units  column  was'  347. 
In  this  case,  the  7  would  be  placed  under  the  units  column 
and  the  34  mentally  added  in  with  the  tens  column. 

29.  Rule. — I.  Write  the  nunibers  so  that  all  the  figures 
of  the  same  order  will  be  in  the  same  column. 

II.  Add  all  the  figures  in  the  right-hand  column ,  or 
column  of  units ,  and  if  their  sum  is  less  than  ten ,  write  it 
underneath.  If  the  sum  is  ten  or  more ,  write  down  the  unit 


1 


ARITHMETIC 


11 


figure  only  and  add  the  figure  or  figures  denoting  the  tens 
in  with  the  next  column. 

III.  Proceed  in  like  manner  with  each  column  until  all  are 
added ,  taking  care  to  write  down  the  entire  sum  of  the  last ,  or 
left-hand ,  column. 

30.  Proof. — There  is  no  complete  proof  possible  of 
addition.  The  usual  method  is  to  repeat  the  operation  in 
the  reverse  order;  that  is,  add  each  column  from  top  to 
bottom.  If  the  same  result  is  obtained  as  by  adding  from 
bottom  to  top,  the  work  is  probably  correct. 

Another  method  is  to  add  the  columns  of  figures  by  the 
method  of  addition  explained  in  example  2,  Art.  27. 


EXAMPLES  FOR  PRACTICE 
Find  the  sum  of: 


(а)  24  +  100  +  365. 

(б)  310  +  985  +  67  +  546. 

(c)  444  +  1,362  +  4,678  +  5,003. 

(d)  7,891  +  338  +  467  +  2,136/ 

(e)  9  +  90  +  900  +  9,000  +  90,000. 

(/)  1,234  +-2,341  +  3,412  +  4,123.  1 

\g)  34,567  +  8,934  +  303+4,212. 


'(a)  489 
(5)  1,908 


Ans.  • 


(c)  11,487 
(if)  10,832 
(e)  99,999 


(f)  11,110 
-(g)  48,016 


1.  What  is  the  sum  of  five  million  three  hundred  seventy  thousand 


four  hundred  seven  plus  seven  hundred  seventy-four  thousand  three 
hundred  forty-four  pluk  five  thousand  three  hundred  ninety-five? 

:r  ^  V»  -  Ans.  6,150,146 


2.  A  contractor  was  employed  1  weeks  in  erecting  a  chimney. 
The  first  week  he  built  30  feet,  the  seeqpd  week  40  feet,  the  third  and 
fourth  weeks  each  50  feet;  what  was  the  height  of  the  chimney? 

Ans.  170  ft. 


3.  A  large  corporation  operates  three  weave  sheds;  the  production 
of  No.  1  weave  room  was  236,450  yards  of  sheeting  for  a  certain  week; 
during  the  same  week,  weave  rooms  Nos.  2  and  3  produehd,  respect¬ 
ively,  125,325  and  133,650  yards  of  fancy  goods;  what  was  the  total 
production  in  yards  for  the  week?  Ans.  495,425  yd. 


4.  A  mill  receives  machinery,  stock,  and  supplies  during  the  first 
week  of  the  month  valued  at  $3,475;  during  the  second  week,  $2,950 
worth;  during  the  third  week,  $4,380  Worth;  and  during  the  rest  of  the 
month,  $4,895  worth;  what  was  the  total  expense  of  these  materials 
for  the  month?  Ans.  $15,700 


12 


ARITHMETIC 


1 


SUBTRACTION 

31.  Subtraction  is  the  process  of  finding  what  part  of 
a  given  quantity  remains  when  a  certain  part  is  taken  from 
it,  and  is  the  reverse  of  the  process  of  addition. 

The  minuend  is  the  given  quantity  that  is  the  larger  of 
the  two  numbers  involved  in  the  process  of  subtraction. 

The  subtrahend  is  the  given  part  that  is  taken  from  the 
minuend  and  thus  is  necessarily  the  smaller  of  the  two 
numbers. 

The  remainder,  or  difference,  is  the  part  of  the  minu¬ 
end  that  is  left  after  the  subtrahend  has  been  taken  from  it. 

32.  The  sign  of  subtraction  is  a  short  horizontal 
line,  thus  —  .  It  is  read  minus,  and  means  less.  Thus, 
8  —  5  is  read  8  minus  5,  and  means  that  5  is  to  be  subtracted , 
or  taken  from  8. 

The  whole  example  would  be  written  8  —  5  =  3,  and  would 
be  read  8  minus  5  equals  3.  In  this  case,  the  number  8  is 
the  minuend,  5  is  the  subtrahend,  and  3  is  the  remainder. 

33.  Only  like  numbers  can  be  subtracted,  as  it  is  evidently 
impossible  to  subtract  feet  from  dollars  or  pounds  from  yards. 

34.  When  one  number  is  to  be  subtracted  from  another, 
the  subtrahend  is  placed  tmder  the  minuend  in  such  a  position 
that  all  the  figures  of  the  same  order  will  be  in  the  same 
column,  or,  in  other  words,  units  must  be  placed  under  units , 
tens  under  tens,  and  so  on. 

After  the  subtrahend  is  placed  under  the  minuend,  a  line 
is  drawn  below  it  and  the  result  of  the  subtraction,  or  the 
remainder ,  placed  beneath  this  line. 

Example. — From  7,849  subtract  5,323. 

Solution. —  minuend  7  8  4  9 

subtrahend  5  3  2  3 


remainder  2  5  2  6  Ans. 


§1 


ARITHMETIC 


13 


Explanation. — Begin  at  the  right-hand,  or  units,  column 
and  subtract  in  succession  each  figure  in  the  subtrahend  from 
the  one  directly  above  it  in  the  minuend,  and  write  the  remain¬ 
ders  below  the  line.  The  result  is  the  entire  remainder. 

35.  It  will  be  noticed  in  the  example  just  given  that  in 
no  case  does  the  value  of  the  subtrahend  figure  of  any  order 
exceed  in  value  the  minuend  figure  of  the  same  order.  If  in 
any  case  the  minuend  figure  of  any  order  (excepting  the  order 
on  the  extreme  left)  has  a  value  less  than  the  corresponding 
subtrahend  figure,  then  another  operation  is  involved,  namely, 
that  of  borrowing.  If  the  minuend  figure  of  the  left-hand 
order  is  of  a  less  value  than  the  subtrahend  figure  of  the  same 
order,  then  subtraction  is  impossible,  since  the  whole  minuend 
is  smaller  than  the  subtrahend,  and  in  arithmetic  a  larger 
value  cannot  be  taken  from  a  smaller. 

Example  1. — From  8,453  take  844. 

Solution. —  minuend  8  4  5  3 

subtrahend  8  4  4 

remahider  7  6  0  9  Ans. 

Explanation. — Begin  at  the  right-hand,  or  units,  column 
to  subtract.  We  cannot  take  4  from  3,  and  must,  therefore, 
borrow  1  from  the  5  in  the  tens  column  and  annex  it  to  the  3 
in  the  units  column.  The  1  ten  =  10  units,  which  added  to 
the  3  in  the  units  column  =  13  units.  4  from  13  =  9,  the 
first,  or  units,  figure  in  the  remainder. 

Since  we  borrowed  1  from  the  5,  only  4  remains;  4  from  4 
=  0,  the  second,  or  tens,  figure.  We  cannot  take  8  from  4, 
and  must,  therefore,  borrow  1  from  8  in  the  thousands  column. 
Since  1  thousand  =  10  hundreds,  10  hundreds  +  4  hundreds 
=  14  hundreds,  and  8  from  14  =  6,  the  third,  or  hundreds, 
figure  in  the  remainder. 

Since  we  borrowed  1  from  8,  only  7  remains,  from  which 
there  is  nothing  to  subtract;  therefore,  7  is  the  next  figure  in 
the  remainder,  or  answer. 

The  operation  of  borrowing  is  performed  by  mentally 
placing  1  before  the  figure  following  the  one  from  which  it  is 
borrowed.  In  the  above  example  the  1  borrowed  from  5  is 


14 


ARITHMETIC 


§1 


placed  before  3,  making  it  13,  from  which  we  subtract  4. 
The  1  borrowed  from  8  is  placed  before  4,  making  14,  from 
which  8  is  taken. 

Example  2. — Find  the  difference  between  10,000  and  8,763. 

Solution. —  minuend  100  00 

subtrahend  8  7  6  3 

remainder  12  3  7  Ans. 

Explanation. — In  the  above  example,  we  borrow  1  from 
the  second  column  and  place  it  before  0,  making  10; 
3  from  10  =  7.  In  the  same  way  we  borrow  1  and  place 
it  before  the  next  cipher,  making  10;  but  as  we  have 
borrowed  1  from  this  column  and  have  taken  it  to  the 
units  column,  only  9  remains  from  which  to  subtract  6; 
6  from  9  =  3.  For  the  same  reason  we  subtract  7  from  9 
and  8  from  9  for  the  next  two  figures,  and  obtain  a  total 
remainder  of  1,237. 

36.  Rule. — I.  Place  the  subtrahend  (or  smaller  number) 
under  the  minuend  {or  larger  number)  so  that  the  figures  of  the 
same  order  will  be  in  the  same  column.  Commencing  at  the 
right,  subtract  each  order  of  the  subtrahend  from  the  correspond- 
ing  order  of  the  minuend  and  write  the  result  in  the  same  order 
of  the  remainder. 

II.  If  any  order  of  the  minuend  has  a  less  vahie  than  the 
corresponding  order  of  the  subtrahend ,  increase  its  value  by  10 
and  subtract;  then  diminish  by  1  the  figure  in  the  next  higher 
order  of  the  minuend  and  subtract. 

37.  There  are  two  methods  of  proving  subtraction,  as 
follows: 

Proof  1. — Add  the  remainder  and  the  subtrahend,  and,  if 
the  work  is  correct,  their  sum  will  equal  the  minuend. 

Example  1. — Prove  that  the  difference  between  1,000  and  987  is  13. 

Solution. —  subtrahend  98  7 

remainder  1 3 

minuend  10  00  Ans. 

This  method  of  proof  depends  on  the  principle  that  sub¬ 
traction  is  the  reverse  of  addition,  and  vice  versa. 


1 


ARITHMETIC 


15 


Proof  2. — Subtract  the  remainder  from  the  minuend  and 
if  the  work  is  correct  the  result  will  equal  the  subtrahend. 

Example  2. — Prove  that  1,265  minus  823  equals  442. 

Solution. —  minuend  12  65 

remainder  4  4  2 

subtrahend  8  23  Ans. 

This  method  of  proof  depends  on  the  principle  that  the 
smaller  of  any  two  numbers  is  equal  to  the  larger  less  the 
difference  of  the  numbers. 


EXAMPLES  FOR  PRACTICE 


Subtract: 


(a)  31  from  58. 

(b)  92  from  99. 

(c)  328  from  649 . 

{d)  837  from  979. 

(e)  345  from  568. 

(/)  799  from  843. 

(g)  876  from  1,123. 

(/;)  7,346  from  9,258. 

(i)  8,675  from  ( 175  +  8,700  +  98) . 

(/)  3,258  from  (2,001  +  7,890). 


f(«) 

27 

-(*) 

7 

(c) 

321 

id) 

142 

\(e) 

223 

(/) 

44 

(g) 

247 

W 

1,912 

(i) 

298 

(/) 

6,633 

Note —In  the  last  two  examples,  it  will  be  noticed  that  a  portion  of  the  example 
is  included  in  parentheses,  thus  (  ).  In  both  cases  this  indicates  that  the  operations 

indicated  within  the  parentheses  are  to  be  performed  first;  and  after  the  value  of  the 
part  within  the  parentheses  is  obtained  as  a  single  number,  the  subtraction  is  per¬ 
formed  as  in  the  preceding  examples. 


i 


16 


ARITHMETIC 


1 


MULTIPLICATION 

38.  Multiplication  is  a  short  process  of  adding  a 
number  to  itself  a  definite  number  of  times. 

The  multiplicand  is  the  number  that  is  added,  or 
multiplied. 

The  multiplier  is  the  number  that  indicates  the  number 
of  times  the  multiplicand  is  to  be  added  to  itself,  or  it  is  the 
number  by  which  we  multiply. 

The  product  is  the  result  obtained  by  multiplication,  or 
the  number  obtained  by  adding  the  multiplicand  to  itself  the 
number  of  times  indicated  by  the  multiplier. 

39.  The  multiplier  always  signifies  a  number  of  times  and 
is  therefore  an  abstract  number. 

The  multiplicand  may  be  either  an  abstract  or  a  concrete 
number. 

The  product  and  the  multiplicand  are  always  like  numbers. 

40.  The  sign  of  multiplication  is  a  small  oblique 
cross,  thus  X,  and  is  placed  between  the  multiplicand  and 
the  multiplier. 

When  the  multiplier  precedes  the  multiplicand  the  sign  of 
multiplication  is  read  times;  thus  6  X  $5  =  $30  is  read  six 
times  five  dollars  equals  thirty  dollars.  In  this  example  the 
figure  6,  being  an  abstract  number,  must  be  the  multiplier. 

When  the  multiplicand  precedes  the  multiplier,  the  sign  of 
multiplication  is  read  multiplied  by;  thus,  $5x6  =  $30  is 
read  five  dollars  multiplied  by  six  equals  thirty  dollars. 

41.  In  the  table  on  the  following  page,  the  product  of 
any  two  numbers  (neither  of  which  exceeds  12)  may  be 
found. 


§1 


ARITHMETIC 


17 


I 

times 

i 

is 

i 

2 

times 

1 

is 

2 

3 

times 

I 

is 

3 

I 

times 

2 

is 

2 

2 

times 

2 

is 

4 

3 

times 

2 

is 

6 

I 

times 

3 

is 

3 

2 

times 

3 

is 

6 

3 

times 

3 

is 

9 

I 

times 

4 

is 

4 

2 

times 

4 

is 

8 

3 

times 

4 

is 

12 

I 

times 

5 

is 

5 

2 

times 

5 

is 

10 

3 

times 

5 

is 

15 

I 

times 

6 

is 

6 

2 

times 

6 

is 

12 

3 

times 

6 

is 

18 

I 

times 

7 

is 

7 

2 

times 

7 

is 

14 

3 

times 

7 

is 

21 

I 

times 

8 

is 

8 

2 

times 

8 

is 

16 

3 

times 

8 

is 

24 

I 

times 

9 

is 

9 

2 

times 

9 

is 

18 

3 

times 

9 

is 

27 

I 

times 

IO 

is 

IO 

2 

times 

IO 

is 

20 

3 

times 

10 

is 

30 

I 

times 

1 1 

is 

1 1 

2 

times 

1 1 

is 

22 

3 

times 

11 

is 

33 

I 

times 

12 

is 

12 

2 

times 

12 

is 

24 

3 

times 

12 

is 

36 

4 

times 

I 

is 

4 

5 

times 

1 

is 

5 

6 

times 

1 

is 

6 

4 

times 

2 

is 

8 

5 

times 

2 

is 

10 

6 

times 

2 

is 

12 

4 

times 

3 

is 

12 

5 

times 

3 

is 

15 

6 

times 

3 

is 

18 

4 

times 

4 

is 

16 

5 

times 

4 

is 

20 

6 

times 

4 

is 

24 

4 

times 

5 

is 

20 

5 

times 

5 

is 

25 

6 

times 

5 

is 

30 

4 

times 

6 

is 

24 

5 

times 

6 

is 

30 

6 

times 

6 

is 

36 

4 

times 

7 

is 

28 

5 

times 

7 

is 

35 

6 

times 

7 

is 

42 

4 

times 

8 

is 

32 

5 

times 

8 

is 

40 

6 

times 

8 

is 

48 

4 

times 

9 

is 

36 

5 

times 

9 

is 

45 

6 

times 

9 

is 

54 

4 

times 

IO 

is 

40 

5 

times 

IO 

is 

50 

6 

times 

10 

is 

60 

4 

times 

1 1 

is 

44 

5 

times 

1 1 

is 

55 

6 

times 

1 1 

is 

66 

4 

times 

12 

is 

48 

5 

times 

12 

is 

60 

6 

times 

12 

is 

72 

7 

times 

I 

is 

7 

8 

times 

1 

is 

8 

9 

times 

1 

is 

9 

7 

times 

2 

fs 

14 

8 

times 

2 

is 

16 

9 

times 

2 

is 

18 

7 

times 

3 

is 

21 

8 

times 

3 

is 

24 

9 

times 

3 

is 

27 

7 

times 

4 

is 

28 

8 

times 

4 

is 

32 

9 

times 

4 

is 

36 

7 

times 

5 

is 

35 

8 

times 

5 

is 

40 

9 

times 

5 

is 

45 

7 

times 

6 

is 

42 

8 

times 

6 

is 

48 

9 

times 

6 

is 

54 

7 

times 

7 

is 

49 

8 

times 

7 

is 

56 

9 

times 

7 

is 

63 

7 

times 

8 

is 

56 

8 

times 

8 

is 

64 

9 

times 

8 

is 

72 

7 

times 

9 

is 

63 

8 

times 

9 

is 

72 

9 

times 

9 

is 

81 

7 

times 

IO 

is 

70 

8 

times 

IO 

is 

80 

9 

times 

10 

is 

90 

7 

times 

ii 

is 

77 

8 

times 

11 

is 

88 

9 

times 

11 

is 

99 

7 

times 

12 

is 

84 

8 

times 

12 

is 

96 

9 

times 

12 

is 

108 

IO 

times 

I 

is 

IO 

11 

times 

1 

is 

11 

12 

times 

1 

is 

12 

IO 

times 

2 

is 

20 

1 1 

times 

2 

is 

22 

12 

times 

2 

is 

24 

IO 

times 

3 

is 

30 

11 

times 

3 

is 

33 

12 

times 

3 

is 

36 

IO 

times 

4 

is 

40 

1 1 

times 

4 

is 

44 

12 

times 

4 

is 

48 

IO 

times 

5 

is 

50 

11 

times 

5 

is 

55 

12 

times 

5 

is 

60 

IO 

times 

6 

is 

60 

11 

times 

6 

is 

66 

12 

times 

6 

is 

72 

IO 

times 

7 

is 

70 

1 1 

times 

7 

is 

77 

12 

times 

7 

is 

84 

IO 

times 

8 

is 

80 

11 

times 

8 

is 

88 

12 

times 

8 

is 

96 

IO 

times 

9 

is 

90 

11 

times 

9 

is 

99 

12 

times 

9 

is 

108 

TO 

times 

IO 

is 

100 

11 

times 

IO 

is 

1 10 

12 

times 

10 

is 

120 

IO 

times 

ii 

is 

I  IO 

11 

times 

1 1 

is 

121 

12 

times 

1 1 

is 

132 

IO 

times 

12 

is 

120 

11 

times 

12 

is 

132 

12 

times 

12 

is 

144 

This  table  should  be  carefully  committed  to  memory. 
Since  0  has  no  value,  the  product  of  0  and  any  number  is  0. 


18 


ARITHMETIC 


1 


42,  To  multiply  a  number  by  one  figure  only: 

Example. — Multiply  425  by  5. 

Solution. —  multiplicand  4  25 

multiplier  5 

product  212  5  Ans. 

Explanation. — For  convenience,  the  multiplier  is  gener¬ 
ally  written  under  the  right-hand  figure  of  the  multiplicand. 
On  looking  in  the  multiplication  table,  we  see  that  5  X  5  is  25. 
Multiplying  the  first  figure  at  the  right  of  the  multiplicand, 
or  5,  by  the  multipler,  5,  it  is  seen  that  5  times  5  units  is 
25  units,  or  2  tens  and  5  units.  Write  the  5  units  in  the  units 
place  in  the  product  and  reserve  the  2  tens  to  add  to  the 
product  of  tens.  Looking  in  the  multiplication  table  again, 
we  see  that  5  X  2  is  10.  Multiplying  the  second  figure  of 
the  multiplicand  by  the  multiplier  5,  we  see  that  5  times 
2  tens  is  10  tens,  and  10  tens  plus  the  2  tens  reserved  are 
12  tens,  or  1  hundred  plus  2  tens.  Write  the  2  tens  in  the 
tens  place,  and  reserve  the  1  hundred  to  add  to  the  product 
of  hundreds.  Again,  we  see  by  the  multiplication  table  that 
5  X  4  is  20.  Multiplying  the  third,  or  last,  figure  of  the 
multiplicand  by  the  multiplier  5,  we  see  that  5  times  4  hun¬ 
dreds  is  20  hundreds,  and  20  hundreds  plus  the  1  hundred 
reserved  are  21  hundreds,  or  2  thousands  and  1  hundred, 
which  we  write  in  the  thousands  and  the  hundreds  places, 
respectively.  Hence,  the  product  is  2,125.  This  result  is 
the  same  as  adding  425  five  times.  Thus, 

425 

425 

425 

425 

425 


sum  2  12  5  Ans. 


§1 


ARITHMETIC 


19 


EXAMPLES  FOR  PRACTICE 
Find  the  product  of: 


(a) 

61,483  X  6. 

(«) 

368,898 

(6) 

12,375  X  5. 

(b) 

61,875 

(c) 

10,426  X  7. 

(c) 

72,982 

(d) 

10,835  X  3. 

Ans.- 

(d) 

32,505 

(e) 

98,376  X  4. 

4 

(e) 

393,504 

(f) 

10,873  X  8. 

( f ) 

86,984 

(g) 

71,543  X  9. 

(g) 

643,887 

(h) 

218,734  X  2. 

(h) 

437,468 

43.  To  multiply  a  number  by  two  or  more  figures: 

Example. — Multiply  475  by  234. 

Solution. —  multiplicand  4  7  5 

multiplier  2  3  4 

19  0  0 
14  2  5 
9  5  0 

product  11115  0  Ans. 

Explanation. — For  convenience,  the  multiplier  is  gener¬ 
ally  written  under  the  multiplicand,  placing  units  under  units, 
tens  under  tens,  etc. 

We  cannot  multiply  by  234  at  one  operation;  we  must, 
therefore,  multiply  by  the  parts  and  then  add  the  partial 
products. 

The  parts  by  which  we  are  to  multiply  are  4  units,  3  tens, 
and  2  hundreds.  4  times  475  =  1,900,  the  first  partial 
product;  3  times  475  =  1,425,  the  second  partial  product ,  the 
right-hand  figure  of  which  is  written  directly  under  the  figure 
multiplied  by ,  or  3;  2  times  475  =  950,  the  third  partial 
product ,  the  right-hand  figure  of  which  is  written  directly 
under  the  figure  multiplied  by,  or  2. 

The  sum  of  these  three  partial  products  is  111,150,  which 
is  the  entire  product. 

44.  Rule.— I.  Write  the  m^dtiplier  under  the  multipli¬ 

cand ,  so  that  units  are  under  units,  tens  under  tens ,  etc. 

II.  Begin  at  the  right  and  multiply  each  figure  of  the  mul¬ 
tiplicand  by  each  successive  figure  of  the  multiplier ,  placing  the 


20  ARITHMETIC  §  1 

right-hand  figure  of  each  partial  product  directly  under  the 
figure  Jised  as  a  multiplier. 

III.  The  sum  of  the  partial  products  will  equal  the  required 
Product. 

45.  Proof. — Review  the  work  carefully;  or  multiply  the 
multiplier  by  the  multiplicand,  afid  if  the  results  agree,  the 
work  is  correct. 

46.  When  there  is  a  cipher  in  the  multiplier ,  multiply 
by  it  the  same  as  with  the  other  figures.  Since  the  product 
of  any  number  and  0  is  0,  the  work  may  be  greatly  short¬ 
ened  by  writing  the  first  cipher  of  the  partial  product,  then 
multiplying  by  the  next  figure  of  the  multiplier  and  writing 
the  partial  product  beside  the  cipher,  as  shown  in  the  follow¬ 
ing  examples: 


(a) 

(■ b ) 

(c) 

(d) 

0 

2 

1  5 

7  0  8 

X 

0 

X  0 

X 

0 

X  0 

0 

Ans. 

0 

Ans. 

0  0  Ans. 

0  0  0 

Ans. 

(e) 

(f) 

(g) 

3 

1  1 

4 

4 

0  0 

8 

3  1  2 

6  4 

2  0 

3 

3  0 

5 

1  0 

0  2 

9 

3  4 

2 

2  0 

0  4 

0 

6  2  5 

2  8 

2  2 

8  0 

12  0  2 

4  0 

312640 

0 

3  2 

1  4 

2 

Ans. 

12  2  2 

4  4 

0 

Ans. 

313265 

2  8  Ans. 

47. 

If 

the 

multiplier 

or 

multiplicand, 

or  both, 

ends  in 

one  or  more  ciphers,  write  the  multiplier  so  that  the  right- 
hand  digit  of  the  multiplier  is  under  the  right-hand  digit  of 
the  multiplicand;  then,  disregarding  the  ciphers  on  the  right 
of  either  multiplier  or  multiplicand,  or  both,  multiply  as 
previously  directed  and  write  as  many  ciphers  on  the  right 
of  the  product  as  there  are  ciphers  on  the  right  of  multiplier 
and  multiplicand.  Thus: 

Example.  — Find  the  product  of:  (a)  78,392  X  248,000;  (b)  7,839,200 
X  248;  (<-)  7,839,200  X  248,000. 


§1 


ARITHMETIC 


21 


Solution. — 

(a) 

7  8  3  9  2 

248000 

627136 

313568 

156784 

194412160  0  0  Ans. 

(c) 

7839200 
248000 
6  2  7  1  3  6 
3  1  3  5  6  8 
156784 


(*) 

7839200 
2  4  8 
627136 
313568 
156784 

1  944121600  Ans. 


1944121600000  Ans. 


EXAMPLES  FOR  PRACTICE 


Find  the  product  of: 


(a) 

13  X  3. 

(*) 

2,679  X  5. 

(c) 

2,893  X  14. 

(d) 

6,734  X  30. 

(e) 

39,465  X  89. 

(/) 

14,674  X  103. 

U) 

367,801  X  2,705. 

w 

567,893  X  1,003. 

'(a) 

39 

(6) 

13,395 

(c) 

40,502 

(d) 

202,020 

(e) 

3,512,385 

(/) 

1,511,422 

Or) 

994,901,705 

(h) 

569,596,679 

1.  A  certain  mill  produces  on  an  average  68,725  yards  of  cloth 
per  day;  what  is  the  production  in  a  year  of  296  working  days? 

Ans.  20,342,600  yd. 

2.  A  reservoir  is  fed  by  two  pumps  each  having  a  capacity  of 

875  gallons  per  hour.  During  a  certain  week  one  pump  was  in 
operation  39  hours  and  the  other  55  hours;  during  the  same  week 
75,000  gallons  of  water  was  drawn  from  the  reservoir;  what  was  the 
excess  of  water  supplied  to  the  reservoir  by  the  pumps  during  the 
week?  Ans.  7,250  gal. 


3.  A  certain  mill  contains  1,830  looms,  each  of  which  produces 
175  yards  of  cloth  per  week;  what  is  the  total  production? 

Ans.  320,250  yd. 


4.  During  a  snow  storm,  snow  fell  to  such  an  extent  that  the  pres¬ 
sure  on  the  roof  of  a  mill  was  4  pounds  per  square  foot.  The  area 
of  the  roof  is  23,625  square  feet;  what  was  the  total  weight  of  snow 
supported  by  the  roof?  Ans.  94,500  lb. 


22 


ARITHMETIC 


1 


DIVISION 

48.  Division  is  a  process  of  finding  how  many  times 
one  number  is  contained  in  another. 

The  number  that  is  divided  is  called  the  dividend. 

The  number  by  which  the  dividend  is  divided  is  called  the 
divisor. 

The  number  that  denotes  the  number  of  times  the  divisor 
is  contained  in  the  dividend  is  called  the  quotient. 

49.  When  the  dividend  does  not  contain  the  divisor  an 
exact  or  even  number  of  times,  the  excess  is  called  the 
remainder.  The  remainder,  being  part  of  the  dividend,  is 
always  of  the  same  kind  as  the  dividend,  and  must  always  be 
less  than  the  divisor. 

50.  If  the  divisor  and  dividend  denote  the  same  kind  of 
units,  then  the  quotient  is  an  abstract  number;  thus,  to  divide 
20  looms  by  5  looms  is  to  find  the  number  of  times  5  looms 
must  be  taken  to  obtain  20  looms,  or  4  times. 

51.  If  the  divisor  is  an  abstract  number,  then  the  quo¬ 
tient  denotes  units  of  the  same  kind  as  the  dividend;  thus,  to 
divide  20  looms  by  4  is  to  find  the  number  of  looms  in  each 
part  when  the  20  looms  are  divided  into  4  equal  parts. 

52.  The  sign  of  division  is  a  short  horizontal  line  with 
a  dot  above  and  a  dot  below  it,  thus  -4-  ,  and  indicates  that 
the  number  preceding  the  sign  is  to  be  divided  by  the  num¬ 
ber  following  it,  the  sign  being  read  divided  by.  Thus,  the 
expression  12  -4-  6  =  2  is  read  12  divided  by  6  equals  2,  or 
G  into  12  equals  2. 

Division  is  also  indicated  by  writing  the  dividend  above  a 
short  horizontal  line  and  the  divisor  below;  thus,  f  =  2,  and 
is  read  8  divided  by  4  equals  2  or  8  over  4  equals  2. 

Division  is  the  reverse  of  multiplication. 


1 


ARITHMETIC 


23 


53.  Dividing  or  multiplying  both  the  divisor  and  divi¬ 
dend  by  the  same  number  does  not  alter  the  value  of  the 
quotient. 

54.  To  divide  wlien  the  divisor  consists  of  but  one 
figure: 

Example. — What  is  the  quotient  of  875  4-  7? 

divisor  dividend  Quotient 

Solution. —  7)875(125  Ans. 

7 

T  7 

1  4 

~3  5 
3  5 

remainder  0 

Explanation. —  7  is  contained  in  8  once.  Place  the  1 

as  the  first,  or  left-hand,  figure  of  the  quotient.  Multiply 
the  divisor  7  by  the  1  of  the  quotient,  and  place  the  prod¬ 
uct  7  under  the  8  in  the  dividend,  and  subtract.  Beside  the 
remainder  1,  bring  down  the  next  figure  of  the  dividend,  in 
this  case  7,  making  17;  7  is  contained  in  17,  2  times.  Write 
the  2  as  the  second  figure  of  the  quotient.  Multiply  the 
divisor  7  by  the  2  in  the  quotient,  and  subtract  the  product 
from  17.  Beside  the  remainder  3,  bring  down  the  next 
figure  of  the  dividend,  in  this  case  5,  making  35.  7  is  con¬ 
tained  in  35,  5  times,  which  is  placed  in  the  quotient.  Mul¬ 
tiplying  the  divisor  by  the  last  figure  of  the  quotient, 
5  times  7  =  35,  which,  subtracted  from  35  under  which  it 
is  placed,  leaves  0.  Therefore,  the  quotient  is  125.  This 
method  is  called  long  division. 

55.  In  sliort  division,  only  the  divisor,  dividend,  and 
quotient  are  written,  the  operations  being  performed  mentally. 

dividend 

divisor  7  )  8  17  35 

quotient  12  5  Ans. 

The  mental  operation  is  as  follows:  7  is  contained  in  8, 
once  and  1  remainder;  imagine  1  to  be  placed  before  7, 
making  17;  7  is  contained  in  17,  2  times  and  3  over;  imagine 


24 


ARITHMETIC 


§1 


3  to  be  placed  before  5,  making  35;  7  is  contained  in  35, 

5  times.  These  partial  quotients,  placed  in  order  as  they 
are  found,  make  the  entire  quotient  125. 

56.  To  divide  when  the  divisor  consists  of  two 
or  more  figures: 

Example. — Divide  2,702,836  by  63. 

divisor  dividend  quotient 

Solution.—  63)2702836(42902  Ans. 

2  5  2 

18  2 
12  6 

5  6  8 
5  6  7 

1  3  6 
12  6 

1  0 

Explanation. — As  63  is  not  contained  in  the  first  two 
figures,  27,  we  must  use  the  first  three  figures,  270.  Now, 
by  trial  we  must  find  how  many  times  63  is  contained  in  270. 

6  is  contained  in  the  first  two  figures  of  270,  4  times.  Place 

the  4  as  the  first,  or  left-hand,  figure  in  the  quotient.  Mul¬ 
tiply  the  divisor,  63,  by  4,  and  subtract  the  product,  252,  from 
270.  The  remainder  is  18,  beside  which  we  write  the  next 
figure  of  the  dividend,  2,  making  182.  Now,  6  is  contained 
in  the  first  two  figures  of  182,  3  times,  but  on  multiplying 
63  by  3,  we  see  that  the  product,  189,  is  too  great,  so  we 
try  2  as  the  second  figure  of  the  quotient.  Multiplying  the 
divisor,  63,  by  2  and  subtracting  the  product,  126,  from  182, 
the  remainder  is  56,  beside  which  we  bring  down  the  next 
figure  of  the  dividend,  making  568.  6  is  contained  in  56 

9  times.  Multiply  the  divisor,  63,  by  9  and  subtraQt  the 
product,  567,  from  568.  The  remainder  is  1,  and  bring¬ 
ing  down  the  next  figure  of  the  dividend,  3,  gives  13.  As  13 
is  smaller  than  63,  we  write  0  in  the  quotient  and  bring  down 
the  next  figure,  6,  making  136.  63  is  contained  in  136, 

2  times,  with  a  remainder  of  10.  Therefore,  42,902e3  is 
the  quotient. 


1 


ARITHMETIC 


25 


57.  The  proper  remainder  is  always  smaller  than  the 
divisor.  If,  while  dividing  at  any  time,  the  remainder  is 
found  to  be  larger  than  the  divisor,  it  indicates  that  the 
quotient  figure  is  not  large  enough  and  it  should  be  increased. 

58.  Rule. — I.  Write  the  divisor  at  the  leit  of  the  divi¬ 

dend ,  with  a  line  between  them. 

II.  Find  how  many  times  the  divisor  is  contained  in  the 
lowest  number  of  the  left-hand  figures  of  the  dividend  that  will 
contain  it,  and  write  the  result  at  the  right  of  the  dividend,  with 
a  line  between,  for  the  first  figure  of  the  quotient. 

III.  Multiply  the  divisor  by  this  quotient  figure;  subtract 
the  product  from  the  figures  of  the  dividend  used  to  obtain  the 
quotient  figure,  and  to  the  remainder  annex  the  next  figure  of 
the  dividend:  Divide  as  before,  and  thus  continue  until  all  the 
figures  of  the  dividend  have  been  used. 

IV.  If,  when  a  figure  is  brought  doivn  from  the  dividend , 
the  number  formed  is  smaller  than  the  divisor,  place  a  cipher  in 
the  quotient  and  bring  dozv?i  the  next  figure  of  the  dividend,  and 
so  ozi  until  the  number  is  laige  enough  to  co?itain  the  divisor. 

V.  If  there  is  finally  a  remainder,  write  it  after  the  quotie7it 
with  the  divisor  underneath. 

VI.  If  at  a7iy  time  the  divisor  multiplied  by  the  quotient 
figure  is  larger  than  the  poi'tion  of  the  dividend  bemg  used,  it 
shows  that  the  qziotient  figure  is  too  large  a?id  should  be 
diminished. 

59.  Proof. — Multiply  the  quotient  by  the  divisor,  and 
to  the  product  add  the  remainder,  if  any,  and  the  result 
obtained  will  equal  the  dividend  if  the  work  is  correct.  Thus, 

divisor  dividend  quotient 

23)1383(60^  Ans. 

1  3  8 


3 

0 

retnainder  3 


26 


ARITHMETIC 


§1 


Proof. —  quotient  6  0 

divisor  2  3 

180 

1  2  0 

13  8  0 

remainder  3 

dividejid  13  8  3 

Or,  divide  the  dividend  by  the  quotient,  and,  if  the  work 
is  correct,  the  result  will  equal  the  original  divisor.  If  there 
is  any  remainder  in  the  original  example  omit  it  when  divi¬ 
ding  the  quotient  into  the  dividend  and  the  same  remainder 
will  occur  in  the  proof.  Thus, 

quotient  dividend  divisor 

60  )  1  3  8  3(  2  3 

1  2  0 

183 

1  8  0 

remainder  3 


EXAMPLES  FOR  PRACTICE 


Solve  the  following  expressions: 

(a)  541,926  4-  7,854. 

(t>)  1,460,883  4-  83. 

( e )  350,096  4-  8. 

{d)  962,251  4-  107. 

(e)  417,789  4-  549. 

(/)  190,850  4-  25. 

(g)  99,999,999  4-  3,333. 

(//)  22,407  4-  462. 


'(«) 

69 

(*) 

17,601 

(C) 

43,762 

id) 

8,993 

(e) 

761 

if) 

7,634 

Or) 

30,003 

l(^) 

48| 

1.  A  mill  purchased  cotton  to  the  value  of  $12,600,  averaging  $42 

per  bale;  how  many  bales  were  purchased?  Ans.  300  bales 

2.  A  certain  chimney  contains  1,120,000  bricks;  five  men  wrere 
employed  in  its  construction,  each  man  laying  2,000  bricks  per  day  on 
an  average;  how  long  did  it  take  to  complete  the  brickwork? 

Ans.  112  da. 


§1 


ARITHMETIC 


27 


FACTORS 

60.  The  factors  of  a  number  are  those  numbers  that, 
when  multiplied  together,  produce  the  number. 

61.  A  prime  number  is  a  number  that  has  no  integral 
factor  except  itself  and  one;  or  in  other  words,  a  prime 
number  is  one  that  cannot  be  divided,  without  a  remainder, 
except  by  itself  and  one. 

The  following  table  includes  all  of  the  prime  numbers  up 
to  and  including  499: 


I 

37 

89 

151 

223 

281 

359 

433 

2 

4i 

97 

157 

227 

283 

367 

439 

3 

43 

IOI 

163 

229 

293 

373 

443 

5 

47 

103 

167 

233 

307 

379 

449 

7 

53 

107 

173 

239 

311 

383 

457 

1 1 

59 

109 

179 

241 

3i3 

389 

461 

13 

6i 

113 

181 

251 

3i7 

397 

463 

1 7 

67 

127 

191 

257 

33i 

401 

467 

19 

7 1 

131 

193 

263 

337 

409 

479 

23 

73 

137 

197 

269 

347 

419 

487 

2y 

79 

139 

199 

271 

349 

421 

491 

3i 

83 

149 

21 1 

277 

353 

43i 

499 

62.  A  prime  factor  is  a  factor  that  is  a  prime  number. 

63.  A  composite  number  is  a  number  that  is  the 

product  of  two  or  more  integral  factors;  thus,  15,  35,  or  117 
are  composite  numbers  since  3x5  =  15,  7x5  =  35,  and 
13  X  9  =  117.  , 

When  referring  to  integral  factors,  the  number  itself  and 
one  are  not  considered. 

A  composite  number  can  have  only  one  set  of  prime 
factors;  thus,  the  number  16  cannot  be  expressed  as  the 
product  of  any  set  of  prime  factors  except  2  X  2  X  2  X  2. 


28 


ARITHMETIC 


1 


It  is  the  product  of  4  X  4  and  of  8x2,  but  in  the  former 
case  both  of  the  factors  are  composite,  and  in  the  latter  case 
one  of  them  is  composite. 

64.  An  even  number  is  a  number  that  is  exactly 
divisible  by  two. 

65.  An  odd  number  is  a  number  that  is  not  divisible 
by  two,  without  a  remainder. 

A  number  that  can  be  divided  by  another,  without  a 
remainder,  is  said  to  be  exactly  divisible,  and  the  divisor  is 
called  an  exact  divisor. 

The  process  of  finding  the  prime  factors  of  a  given  number 
is  one  of  repeated  divisions,  using  for  divisors  the  smallest 
prime  numbers  that  are  exactly  divisible  into  the  number, 
or  into  the  quotients  resulting  from  previous  divisions. 

66.  To  find  the  prime  factors  of  a  number: 

Example. — Find  the  prime  factors  of  576. 

Solution. —  2)576 

2  )  2  8  8 
2)144 
2  ) _ 72 

2  ) _ 36 

2  ) _ 18 

3  ) _ 9 

3 

That  is:  2  X  2  X  2  X  2  X  2  X  2  X  3  X  3  =  576.  Ans. 

Explanation. — In  this  example  the  division  is  accom¬ 
plished  by  the  short  method  of  division,  using  the  smallest 
prime  number  possible  for  each  divisor;  these  divisors  and 
the  quotient  of  the  last  division  are  the  prime  factors  of  the 
number  576,  since  they  are  all  prime  numbers,  and  when 
multiplied  together  produce  the  number  576. 

Rule. — Divide  the  given  number  by  the  smallest  prime  number 
that  will  exactly  divide  it.  If  the  quotient  is  a  composite  number , 
proceed  in  the  same  manner;  and  continue  until  a  quotient  is 


1 


ARITHMETIC 


29 


obtained  that  is  a  prime  number.  The  several  divisors  and  the 
last  quotient  are  the  prime  factors  required. 

If  no  prime  factor  is  found  before  the  quotient  becomes 
equal  to,  or  less  than,  the  divisor,  the  number  is  a  prime 
number. 


EXAMPLES  FOR  PRACTICE 
Find  the  prime  factors  of  the  following  numbers: 


( a )  24. 

'(«)  2,  2,  2,  3 

CO 

00 

(b)  2,  19 

(c)  108. 

(c)  2,  2,  3,  3,  3 

(. d )  2,160. 

.  AnsJ 

(d)  2,  2,  2,  2,  3,  3,  3,  5 

(e)  4,224. 

(e)  2,2,2,2,2,2,2,3,11 

(/)  36,960. 

(f)  2,2,2,2,2,3,5,7,11 

(g)  59,049. 

(g)  3,  3,  3,  3,  3,  3,  3,  3,  3,  3 

(h)  11,760. 

[(h)  2,  2,  2,  2,  3,  5,  7,  7 

GREATEST  COMMON  DIVISOR 

67.  A  common  divisor  of  two  or  more  numbers  is  any 
divisor  that  will  exactly  divide  them;  thus,  4  is  a  common 
divisor  of  8,  12,  16,  and  20,  because 

8-4  =  2 
12  -  4  =  3 
16  -  4  =  4 
20  -  4  =  5 

68.  To  find  a  common  divisor  of  two  or  more 
numbers: 

Rule. — Resolve  the  given  numbers  into  two  factors ,  one  of 
which  is  common  to  all;  this  common  factor  is  a  common  divisor 
of  the  given  numbers. 

69.  The  greatest  common  divisor  of  two  or  more 
numbers  is  the  greatest  number  that  will  divide  each  of  them 
without  a  remainder. 

The  divisors  of  36  are  2,  3,  4,  6,  9,  12,  and  18,  and  the 
divisors  of  24  are  2,  3,  4,  6,  8,  and  12.  The  only  common 
divisors  of  36  and  24  are  2,  3,  4,  6,  and  12,  of  which  12  is  the 
greatest;  therefore,  12  is  the  greatest  common  divisor  of  36 
and  24. 


30 


ARITHMETIC 


§1 


If  two  integral  numbers  have  no  common  divisor,  they  are 
said  to  be  prime  to  each  other;  thus,  9  and  16  are  prime  to 
each  other,  although  both  are  composite  numbers. 

The  letters  G.  C.  D.  stand  for  greatest  common  divisor, 
or  in  some  instances,  where  the  term  greatest  common 
measure  is  used,  the  abbreviation  G.  C.  M.  is  adopted. 

In  finding  the  greatest  common  divisor  of  two  or  more 
numbers,  first  reduce  the  numbers  to  their  prime  factors,  and 
then  find  the  product  of  those  factors  that  are  common  to  all 
the  numbers. 

Example. — What  is  the  G.  C.  D.  of  72  and  168? 


Solution. — 

2)72 

2)168 

2)36 

2  )  8  4 

2)18 

2)  4  2 

3  )  9 

3  )  2  1 

3 

7 

2  is  a  common  divisor  of  72  and  168  three  times,  and  3  is  a 
common  divisor  once,  so  the  greatest  common  divisor  of  72  and  168 
is  2  X  2  X  2  X  3  =  24.  Ans. 

70.  To  find  the  greatest  common  divisor  of  two 
or  more  numbers: 

Rule. — Resolve  the  given  numbers  i?ito  their  prime  factors; 
the  product  of  those  factors  that  are  common  to  all  the  given 
numbers  is  the  greatest  common  divisor. 


EXAMPLES  FOR  PRACTICE 


Find  the  G.  C.  D.  of: 

(a)  45  and  75. 

(b)  120  and  168. 

(c)  60  and  220. 

{d)  630  and  2,730. 

( e )  63,  171,  and  207. 

(/)  1,890,  4,410,  and  10,710. 


(a) 

15 

(b) 

24 

(c) 

20 

id) 

210 

(<?) 

9 

(f) 

630 

1 


ARITHMETIC 


31 


LEAST  COMMON  MULTIPLE 

71.  A  multiple  is  a  number  that  can  be  exactly  divided 
by  an  integer,  and  is  said  to  be  a  multiple  of  that  integer, 
thus:  24  is  a  multiple  of  8,  since  it  can  be  divided  exactly  by  8. 

72.  A  common  multiple  of  two  or  more  numbers  is 
a  number  that  can  be  exactly  divided  by  each  of  them;  thus, 
24  is  a  common  multiple  of  2,  3,  4,  6,  8,  and  12. 

73.  The  least  common  multiple  of  two  or  more  num¬ 
bers  is  the  least  number  that  can  be  exactly  divided  by  each 
of  them. 

74.  A  multiple  of  a  number  contains  all  of  the  prime 
factors  of  that  number,  and  a  common  multiple  of  two  or 
more  numbers  contains  all  the  prime  factors  of  all  the 
numbers. 

The  least  common  multiple  of  two  or  more  numbers  is  the 
least  number  that  will  contain  all  the  prime  factors  of  those 
numbers  and  no  others;  therefore,  the  least  common  multiple 
will  have  each  prime  factor,  taken  only  the  greatest  number 
of  times  it  is  contained  in  any  of  the  several  numbers. 

The  letters  L.  C.  M.  stand  for  least  common  multiple. 

75.  To  find  the  least  common  multiple,  arrange  the 
figures  in  a  horizontal  line,  and  divide  them  by  the  smallest 
prime  factors  possible  that  will  divide  two  or  more  of  the 
numbers,  then  find  the  product  of  the  factors  and  the  quo¬ 
tients  remaining. 

Example. — Find  the  L.  C.  M.  of  48,  60,  36,  and  13. 

Solution. — Arranging  the  numbers  as  stated  in  the  above  clause. 

2  )  48,  60,  36,  13 

2  )  24  .  30,  18,  13 

3  )  12,  15,  9,  13 

4,  5,  3,  13 

2X2X3X4X5X3X  13  =  9,360.  Ans. 

Explanation. — In  this  example  all  of  the  numbers  are 
divided  by  the  smallest  prime  number,  which  will  exactly 


32 


ARITHMETIC 


§1 

divide  into  two  or  more  of  them.  This  process  is  then  con¬ 
tinued  with  the  quotients  thus  obtained,  until  quotients  are 
found  that  are  prime  to  each  other.  The  product  of  these 
quotients,  and  the  prime  factors  used  to  obtain  them,  is  equal 
to  9,360,  which  is  the  least  common  multiple  of  48,  60,  36, 
and  13. 

76.  To  find  tlie  least  common  multiple  of  two  or  * 
more  numbers. 

Rule. — Having  arra?iged  the  numbers  in  a  horizontal  line , 
divide  them  by  the  smallest  Prime  number  that  will  divide  two 
or  more  of  them  exactly.  Write  the  quotients  and  the  members 
that  cannot  be  divided  without  a  remainder  miderneath.  Proceed , 
in  like  manner ,  until  no  prime  number  will  divide  more  tha?i  one 
of  the  numbers.  The  product  of  the  divisor,  and  the  numbers  of 
the  last  line ,  is  the  least  common  multiple  of  the  numbers. 


EXAMPLES  FOR  PRACTICE 


Find  the  L.  C.  M.  of 


00 

7,  8 

i,  and  18. 

(b) 

37, 

74,  and  3. 

(c) 

10, 

15,  25,  and  80. 

(d) 

69, 

23,  115,  and  4. 

O’) 

10, 

40,  30,  and  300. 

( f ) 

459 

,  153,  34,  and  2. 

Or) 

22, 

28,  14,  7,  112,  and  36. 

(h) 

9,  8 

I,  12,  18,  24,  and  36. 

'(a) 

126 

(b) 

222 

(c) 

1,200 

(d) 

1,380 

(e) 

600 

(0 

918 

Or) 

11,088 

[(h) 

72 

§1 


ARITHMETIC 


33 


CANCELATION 

77.  Cancelation  is  a  method  of  shortening  division  by 
removing,  or  canceling,  equal  factors  from  the  divisor  and 
dividend. 

The  process  of  cancelation  depends  on  the  principle  that 
dividing  the  dividend  and  the  divisor  by  the  same  number 
does  not  alter  the  value  of  the  quotient. 


78.  Cancelation  is  employed  when  multiplication  is 
involved  with  division;  that  is,  when  certain  numbers 
have  to  be  multiplied  together  and  then  divided  by  one  or 
more  numbers.  By  this  means  long  examples  are  greatly 
simplified. 


Example. — Divide  21  X  36  X  32  by  9  X  7  X  16. 


Solution. — 


3  4  2 

21  X  36  x  32 

$  x  7  X  16 

1  l  l 


Ans. 


Explanation. — Both  9  and  36  can  be  divided  by  9.  Cross 
out  these  two  numbers  and  write  the  result  of  the  division 
over  or  under  the  numbers  divided,  which  will  give  the 
following: 

4 

21  x  n  X  32 
0X7X16 
1 


Next  notice  if  there  are  any  more  numbers  that  can  be 
dealt  with  in  the  same  way.  7  and  21  can  both  be  divided 
by  7.  Cross  out  these  two  numbers  and  write  the  results  as 
before,  when  the  following  will  be  obtained: 

3  4 

UxUX  32  _ 

0  X  7  X  16 

1  1 


34 


ARITHMETIC 


§1 


Continuing  with  this  process  it  will  be  seen  that  16  and  32 
can  be  divided  by  16  and,  consequently,  will  give: 

3  4  2 

UxUX U . 

0  X  t  X  U 

1  1  1 

Since  there  are  no  two  remaining  numbers  (one  in  the  divi¬ 
dend  and  the  other  in  the  divisor)  divisible  by  the  same 
number  except  1  without  a  remainder,  it  is  impossible  to 
cancel  further. 

Multiply  all  the  remaining  numbers  in  the  dividend 
together  and  divide  their  product  by  the  product  of  all  the 
remaining  numbers  in  the  divisor.  The  result  will  be  the 
quotient. 

The  product  of  all  the  remaining  numbers  in  the  dividend 
equals  3x4x2  =  24.  The  product  of  all  the  remaining 
numbers  in  the  divisor  equals  1  X  1  X  1  =  1. 

Hence, 

3  4  2 

UxUx  U  =  3  X  4  X  2  =  24  =  24  Ang 

$Xt  XW  lXlXl  1 
1  1  1 


79.  If,  during  the  process  of  cancelation,  a  number  is 
divided  by  itself,  it  is  not  customary  to  write  down  the 
quotient  1,  but  to  proceed  as  in  the  following  example: 


Example. — Divide  27  X  33  X  15  by  11  X  9  X  5. 


Solution. — 


3  3  3 

n  X  33  X  15  _ 
n  x  d  x  $ 


Ans. 


80.  The  numbers  resulting  from  dividing  a  number  in 
the  dividend  and  a  number  in  the  divisor  by  the  same  num¬ 
ber  may  also  be  canceled  in  like  manner,  as  shown  in  the 
following  example: 

Example.— Divide  4  X 12  X  24  X  20  by  8  X  30  X  2  X  4. 

2 

6  6  2 

i  x  n  x  n  x  m  ... ,»  . 

8x3 0X2X4  12'  Ans' 

%  3 


Solution. — 


§1 


ARITHMETIC 


35 


81.  If  all  the  numbers  in  both  the  dividend  and  divisor 
should  cancel  down  to  1,  the  answer  would  not  be  0,  but 
would  be  1.  This  point  is  clearly  shown  in  the  following 
example,  which  is  carried  through  every  process: 

1  1 

Mil 

Ux$x$xU  =  l  x  l  x  l  x  i  =  l  =  j 
U  x  \%  x  0  x  4  l  x  l  x  i  x  l  l 

2  12  1 

l  l 

82.  Rule. — I.  Cancel  the  factor  or  factors  common  to  the 
dividend  and  divisor. 

II.  Divide  the  product  of  the  factors  remaining  in  the  divi¬ 
dend  by  the  product  of  the  factors  remaining  in  the  divisor. 


EXAMPLES  FOR  PRACTICE 


Solve  the  following  expressions: 


(a) 

(b) 

O 

(d) 

(e) 
(/) 

(g) 

(h) 


8  X  4  =  ? 

2X2 

6  X  9  X  12  =  ? 
9X3X6 
10  X  15  X  25  =  ? 
25  X  5  X  5 
18  X  45  X  54  _  ? 
27  X  9  X  3 
28  X  35  X  63 
14X21 

54  X  72  X  78  _  ? 
12  X  18  X  6 
8  X  12  X  96  =  ? 

48  X  2  X  12 

49  X  35  X  63  X  72 
14  X  21  X  42 


(a) 

00 

(b) 

4 

C) 

6 

(d) 

60 

(e) 

210 

( f ) 

234 

(g) 

8 

(h) 

630 

>  • 


ARITHMETIC 

(PART  2) 


FRACTIONS 

1.  A  fraction  is  one  or  more  parts  of  a  unit.  One-half , 
three-fourths,  two-fifths  are  fractions. 

If  a  weave  room  contains  1,000  looms  and  has  500  of  this 
number  on  fancy  work  and  the  other  500  on  plain  work,  it 
might  be  stated  that  one-half  of  the  looms  were  on  one  class 
of  work  and  one-half  on  another  class  of  work.  Instead  of 
writing  out  the  expression  one-half  in  words,  it  could  be 
shown  by  means  of  figures,  thus  i,  and  would  be  read  one- 
half.  When  any  part  of  a  number  or  unit  is  expressed  in 
this  manner,  it  is  said  to  be  expressed  fractionally ,  and  the 
expression  (in  this  case  2)  is  known  as  a  fraction. 

2.  Two  numbers  are  required  to  express  a  fraction;  one 
is  called  the  numerator,  and  the  other,  the  denominator. 

The  numerator  is  placed  above  the  denominator,  with  a 
line  between  them,  as  f.  Here  3  is  the  denominator,  and 
shows  into  how  many  equal  parts  the  unit,  or  o?ie,  is  divided. 
The  numerator  2  shows  how  many  of  these  equal  parts  are 
taken  or  considered.  The  denominator  also  indicates  the 
names  of  the  parts. 

2  is  read  one-half 
I  is  read  three-fourths 
f  is  read  three-eighths 
te  is  read  five-sixteenths 

is  read  twenty-nine  forty-sevenths 

For  notice  of  copyright,  see  page  immediately  following  the  title  page 

22 


2 


ARITHMETIC 


§2 


3.  In  the  expression  “f  of  an  apple,”  the  denominator  4 
shows  that  the  apple  is  to  be  (or  has  been)  cut  into  4  equal 
Parts ,  and  the  numerator  3  shows  that  three  of  these  parts ,  or 
fourths ,  have  been  taken  or  considered. 

If  each  of  the  parts ,  or  fourths ,  of  the  apple  were  cut  into 
two  equal  pieces ,  there  would  then  be  twice  as  many  pieces  as 
before,  or  4  X  2  =  8  pieces  in  all;  one  of  these  pieces  would 
be  called  one-eighth,  and  would  be  expressed  in  figures  as  i. 
Three  of  these  pieces  would  be  called  three-eighths,  and 
written  f.  The  words  three-fourths,  three-eighths,  five-six¬ 
teenths,  etc.  are  abbreviations  of  three  one-fourths,  three 
one-eighths,  five  one-sixteenths,  etc.  It  is  evident  that  the 
larger  the  denominator,  the  greater  is  the  number  of  parts 
into  which  anything  is  divided;  consequently,  the  parts 
themselves  are  smaller,  and  the  value  of  the  fraction  is  less 
for  the  same  number  of  parts  taken.  In  other  words,  9,  for 
example,  is  smaller  than  i,  because  if  an  object  be  divided 
into  9  parts,  the  parts  are  smaller  than  if  the  same  object 
had  been  divided  into  8  parts;  and,  since  i  is  smaller  than  Jr, 
it  is  clear  that  7  one-ninths  is  a  smaller  amount  than  7  one- 
eighths.  Hence,  also,  I  is  less  than  I. 

The  line  between  the  numerator  and  denominator  means 
divided  by,  or  -f- .  t  is  equivalent  to  3  -h  4.  1  is  equivalent 

to  5  -r-  8. 

4.  The  value  of  a  fraction  is  equal  to  the  numerator 
divided  by  the  denominator,  as  4  =  2,  f  =  3. 

The  value  of  a  fraction  whose  numerator  and  denominator 
are  equal  is  1. 

7,  or  four-fourths  =  1 
I,  or  eight-eighths  =  1 
ti,  or  sixty-four  sixty-fourths  =  1 

5.  The  numerator  and  denominator  of  a  fraction  are 
called  the  terms  of  a  fraction. 

6.  A  proper  fraction  is  a  fraction  whose  numerator  is 
less  than  its  denominator.  Its  value  is  less  than  1.  i,  t,  9 
are  all  proper  fractions,  since  the  number  above  the  line,  or 


2 


ARITHMETIC 


3 


the  numerator,  in  each  case  is  less  than  the  number  below 
the  line,  or  the  denominator. 


7.  An  improper  fraction  is  a  fraction  whose  numerator 

is  egjial  to ,  ox  greater  than ,  its  denominator.  Its  value  is  one 
or  more  than  one.  i  are  -all  improper  fractions,  since 

in  each  case  the  numerator  is  equal  to,  or  greater  than,  the 
denominator. 

8.  A  simple  fraction  is  a  fraction  with  but  one  number 
for  its  numerator  and  one  number  for  its  denominator.  Such 
a  fraction  may  be  either  proper  or  improper.  All  the  illus¬ 
trations  so  far  given  are  simple  fractions. 

9.  A  mixed  number  is  a  whole  number  and  a  fraction 
combined.  8|  (read  eight  and  three-fourths)  is  a  mixed 
number,  as  it  is  composed  of  the  whole  number  8  and  the 
fraction  f . 


10.  A  complex  fraction  is  one  that  has  a  fraction  or  a 
mixed  number  for  either  its  numerator  or  denominator,  or 
for  both. 


— 6,  and  ^  are  examples  of  complex  fractions. 
24  5| 


11.  To  read  a  simple  fraction,  read  the  numerator  as 
though  it  were  a  whole  number  and  then  read  the  denomi¬ 
nator  adding  th  or  ths  to  the  name  of  this  figure,  with  the 
few  exceptions  noted  in  the  following: 

\  is  read  one-half 
-iV  is  read  one-twelfth 
2 h  is  read  one  twenty-second 

All  other  denominators  of  fractions  ending  with  a  2 
(except  those  ending  with  12)  are  read  similar  to  the  last 
example  given  above. 

i  is  read  one-third 
iV  is  read  one-thirteenth 
2V  is  read  one  twenty-third 

All  other  denominators  of  fractions  ending  with  a  3 
(except  those  ending  with  13)  are  read  similar  to  the  last 
example  given  above. 


4 


ARITHMETIC 


2 


i  is  read  one-fifth 
tV  is  read  one-fifteenth 
2V  is  read  one  twenty-fifth 

All  other  denominators  of  fractions  ending  with  a  5 
(except  those  ending  with  15)  are  read  similar  to  the  last 
example  given  above. 

-To  is  read  one-tenth 
2V  is  read  one-twentieth 

All  other  denominators  of  fractions  ending  in  ty  are  spelled 
similar  to  the  last  example. 

The  fraction  4  is  read  one-fourth  and  also  one-quarter. 

All  denominators  of  fractions  having  two  figures  or  more 
and  ending  with  a  4  are  read  by  adding  th  or  ths  to  the  name 
of  the  figure.  _ 


REDUCTION  OF  FRACTIONS 

12.  The  process  of  changing  the  terms  of  a  fraction 
without  changing  the  value  of  the  fraction  is  known  as 
reduction. 

13.  Multiplyivg  both  the  numerator  and  denominator  of 
a  fraction  by  the  same  number  does  not  change  the  value  of  the 
fraction.  Thus,  multiplying  both  terms  of  the  fraction  4  by  2, 

3  X  2  =  6 

4X2  8 

The  value  is  not  changed,  since  f  =  f.  Suppose  that 
an  object,  say  an  apple,  is  divided  into  8  equal  parts.  If 
these  parts  be  arranged  in  4  piles,  each  containing  2  parts, 
it  is  evident  that  each  pile  will  be  composed  of  the  same 
amount  of  the  entire  apple  as  would  have  been  the  case 
had  the  apple  been  originally  cut  into  4  equal  parts.  Now, 
if  one  of  these  piles  (containing  2  parts)  be  removed, 
there  will  be  3  piles  left,  each  containing  2  equal  parts, 
or  6  equal  parts  in  all;  i.  e.,  six-eighths.  But,  since  one 
pile,  or  one-quarter,  was  removed,  there  are  three-quarters 
left.  Hence,  4=1.  The  same  course  of  reasoning  may 
be  applied  to  any  similar  case.  Therefore,  multiplying 


2 


ARITHMETIC 


5 


both  terms  of  a  fraction  by  the  same  number  does  not  alter 
its  value. 

14.  Dividing  both  the  numerator  and  denominator  of  a 
fraction  by  the  same  number  does  not  change  the  value  of  the 
fraction.  Thus,  dividing  both  terms  of  the  fraction  i%-  by  2, 

_8_  -7-  2  _  4 
10  -s-  2  5 

That  =  f  is  readily  seen  from  the  explanation  given  in 
Art.  13;  for,  multiplying  both  terms  of  the  fraction  1  by  2, 

4  X  2  _  8 
5x2  10 

and,  if  i  =  must  equal  1.  Hence,  dividing  both  terms 

of  a  fraction  by  the  same  number  does  not  alter  its  value. 

15.  A  fraction  is  reduced  to  lower  terms  by  dividing  both 
terms  by  the  same  number. 

A  fraction  is  reduced  to  its  lowest  terms  when  its  numer¬ 
ator  and  denominator  cannot  both  be  divided  by  the  same 

number,  except  1,  without  a  remainder;  for  example,  f,  t, 

1  1  _8_ 

2  4,  15. 

16.  Suppose  that  it  is  desired  to  reduce  the  fraction  fo  to 
its  lowest  terms;  that  is,  to  have  the  smallest  possible  numbers 
for  the  numerator  and  denominator. 

Both  numerator  and  denominator  can  be  divided  by  2  and 
the  result  is  if.  Both  the  numerator  and  denominator  of 
this  new  fraction,  if,  cannot  be  divided  by  2  without  a 
remainder,  but  both  can  be  divided  by  3,  giving  the  result  f. 

The  fraction  f  cannot  be  reduced  any  lower,  since  there 
is  no  number  greater  than  1  that  will  exactly  divide  both 
terms  of  the  fraction.  Therefore,  f  is  the  lowest  terms  in 
which  the  fraction  ff  can  be  expressed. 

17.  To  reduce  a  fraction  to  its  lowest  terms: 

Rule. — Divide  both  the  numerator  and  denominator  by  any 
number  greater  than  1  that  will  divide  into  both  without  a 
remainder,  a?id  continue  this  operation  until  there  is  no  number 
greater  than  1  that  will  exactly  divide  both  terms. 


6 


ARITHMETIC 


§2 


EXAMPLES  FOR  PRACTICE 


Reduce  the  following  fractions  to  their  lowest  terms: 


(a)  fi 

( b )  wi. 

(c)  A. 

(d)  n. 


Ans.- 


(a) 

( b ) 
{c) 
(d) 


1_ 

2 

X 

8 

3. 

8 

69 
7  O 


(*) 


2  16 
1  2  9  6* 


(e)  i 


(0 


94 

5  64* 


l(/) 


JL 

6 


18.  A  fraction  is  reduced  to  higher  terms  by  multiplying 
both  the  numerator  and  denominator  by  the  same  number. 

Thus,  if  both  terms  of  the  fraction  I  are  multiplied  by  2, 
the  result  is  the  fraction  f,  which  is  of  the  same  value  as  i. 

19.  To  reduce  a  fraction  to  an  equal  fraction  hav¬ 
ing  a  given  denominator: 

Example. — Reduce  f  to  an  equal  fraction  having96 fora  denominator. 

Solution. — Both  the  numerator  and  the  denominator  must  be  multi¬ 
plied  by  the  same  number  in  order  not  to  change  the  value  of  the  frac¬ 
tion.  The  denominator  must  be  multiplied  by  some  number  which 
will,  in  this  case,  make  the  product  96;  this  number  is  evidently  96  -f-  8 

7  V  12  84 

=  12,  since  8  X  12  =  96.  Hence,  -  x  ^  =  95-  Ans- 

Rule. — Divide  the  denominator  of  the  nezv  fraction  by  the 
denominator  of  the  given  fraction  and  multiply  both  terms  of 
the  fraction  by  this  result. 

20.  To  reduce  a  whole  number  to  an  improper 
fraction: 

Example. — How  many  fifths  in  7? 

Solution.— Since  there  are  5  fifths  in  1  (f  =  1),  in  7  there  will  be 
7X5  fifths,  or  35  fifths;  i.  e.,  7xf  =  sgL.  Ans. 

Rule. — Multiply  the  whole  number  by  the  denominator  of 
the  desired  fraction  and  use  this  result  as  the  numerator. 

21.  To  reduce  a  mixed  number  to  an  improper 
fraction: 

Example. — Reduce  4|  to  eighths. 

Solution. — Since  there  are  8  eighths  in  1  (f  =  1),  in  4  there  will  be 
4X8  eighths,  or  32  eighths,  and  32  eighths  plus  5  eighths  equals 
37  eighths,  or  Therefore,  4|  =  nr.  Ans. 


§2 


ARITHMETIC 


7 


Rule. — Multiply  the  whole  number  by  the  denominator  of  the 
fraction,  and  to  this  result  add  the  numerator  of  the  fraction. 
Use  the  result  tlms  obtained  for  the  numerator  of  the  improper 
fraction,  and  for  the  denominator  take  the  denominator  of  the 
fraction  in  the  mixed  number. 

22.  To  reduce  an  improper  fraction  to  a  whole  or 
a  mixed  number: 

Example  1. — Reduce  ^  to  a  whole  or  a  mixed  number. 

Solution. — Since  one  unit  contains  6  sixths,  there  areas  many  units 
in  or  48  sixths,  as  6  is  contained  times  in  48,  or  8.  Therefore,  the 
whole  number  8  is  equal  to  the  fraction  Ans. 

Example  2. — Reduce  x  to  a  whole  or  mixed  number. 

Solution. — Since  in  one  unit  there  are  8  eighths,  the  number  of 
units  in  ^ ,  or  21  eighths,  must  equal  21  eighths  divided  by  8  or  2  plus 
a  remainder  of  5  eighths.  Therefore,  ^  =  2  +  f  =  2f.  Ans. 

Rule. — Divide  the  numerator  by  the  denominator.  If  there 
is  a  remainder  write  it  over  the  denominator  of  the  fraction 
and  place  the  fraction  thus  formed  with  the  whole  number, 
making  the  result  a  mixed  number. 


EXAMPLES  FOR  PRACTICE 

(a)  Reduce  §  to  a  fraction  having  12  for  a  denominator.  Ans. 

(b)  Reduce  xc  to  a  fraction  having  90  fora  denominator.  Ans.  iH> 
{, c )  Reduce  ^  to  a  fraction  having  96  as  a  denominator.  Ans.  trt 

(d)  Change  6  to  an  improper  fraction  having  24  as  its  denominator. 

Ans.  ^ 

( e )  Change  12  to  an  improper  fraction  having  5  as  its  denominator. 

Ans.  ^ 

(/)  Reduce  14  to  an  improper  fraction  with  3  as  its  denominator. 


Ans.  ^ 

(e) 

Reduce  to  an  improper  fraction. 

Ans.  ff 

(h 

Reduce  12f  to  an  improper  fraction. 

Ans. 

( i ) 

Reduce  4}  to  an  improper  fraction. 

Ans.  ^ 

Change  the  following  to  whole  or  mixed  numbers: 

(/) 

25 

6  • 

Ans.  5 

(h 

4  7 

7  • 

Ans.  6f 

(l) 

1  9 

4  • 

Ans.  4\ 

(m) 

2  7 

3  • 

Ans.  9 

8 


ARITHMETIC 


§2 


23.  A  common  denominator  of  two  or  more  fractions 
is  a  number  that  will  contain  (i.  e.,  that  may  be  divided  by) 
the  denominator  of  each  of  the  given  fractions  without  a 
remainder.  The  least  common  denominator  is  the  least 
number  that  will  contain  each  denominator  of  the  given  frac¬ 
tions  without  a  remainder. 

24.  To  find  the  least  common  denominators 

Example  1. — Find  the  least  common  denominator  of  1,  and  re. 

Solution. — First  place  the  denominators  in  a  row,  separated  by 
commas. 

2)  4,  3,  9,  16 

2  )  2,  3,  9,  8 

3)  1,  3,  9,  4 

3)  1,  1,  3,  4 

4)  1,  1,  1,  4 
1,  1,  1,  1 

2X2X3X3X4  =  144,  the  least  common  denominator.  Ans. 

Explanation. — Divide  each  of  the  numbers  by  some 
prime  number  that  will  divide  at  least  two  of  them  without  a 
remainder  (if  possible) ,  bringing  down  to  the  row  below  those 
denominators  that  will  not  contain  the  divisor  without  a 
remainder.  Dividing  each  of  the  numbers  by  2,  the  second 
row  becomes  2,  3,  9,  8,  since  2  will  not  divide  3  and  9  without 
a  remainder.  Dividing  again  by  2,  the  result  is  1,  3,  9,  4. 
Dividing  the  third  row  by  3,  the  result  is  1,  1,  3,  4.  So  con¬ 
tinue  until  the  last  row  contains  only  l’s.  The  product  of 
all  the  divisors,  or  2  X  2  X  3  X  3  X  4,  is  144,  and  is  the  least 
common  denominator. 

Example  2. — Find  the  least  common  denominator  of  £,  •&,  A- 


Solution. — 

3)9, 

12, 

18 

3)3, 

4, 

6 

2)1, 

4, 

2 

2)1, 

2, 

1 

1,  1,  1 

3  X  3  X  2  X  2  =  36.  Ans. 


25.  To  reduce  two  or  more  fractions  to  fractions 
having  a  common  denominator: 


§2 


ARITHMETIC 


9 


Example. — Reduce  f,  f,  and  £  to  fractions  having  a  common 
denominator. 

Solution. — The  common  denominator  is  a  number  that  will  contain 
3,  4,  and  2.  The  least  common  denominator  is  12,  because  it  is  the 
smallest  number  that  can  be  divided  by  3,  4,  and  2  without  a 
remainder. 

2  _8  3  _  _9  1  _  _6 

3  "  12’  4  ”  12’  2  ~  12 

Reducing  f  (see  Art.  19),  3  is  contained  in  12,  4  times.  By  multi- 

2  4  8 

plying  both  numerator  and  denominator  of  f  by  4,  we  find  -  X  =  ys- 
In  the  same  way,  we  find  f  =  A,  and  |  =  A-  Ans. 

Rule. — For  each  fraction  to  be  reduced,  divide  the  common 
detiominator  by  the  denominator  of  the  fraction ,  and  multiply 
both  terms  ot  the  fraction  by  the  quotient. 


EXAMPLES  FOR  PRACTICE 

Reduce  the  following  to  fractions  having  the  least  common  denom 
inator: 


(a) 

'•fit  to  and  + 

(a)  A, 

8  16  12 
24)  24»  24 

(b) 

H  and  A  • 

(b)  «, 

8 

2  4 

(0 

t>  vr>  and  A" ■ 

Ans.- 

(c)  «, 

10  8 

4  2)  42 

(d) 

■§■>  ts  1  and  5+ 

id)  f§, 

28  12  24 

60)  60)  60 

(*) 

4|,  3-f,  and  5*. 

(p\  4iO  Ol_2  C_3_ 

\C)  ^16)  ^16»  ^16 

(0 

-Z-  3  pn/1  4*. 

16)  4  >  «*UU  24  • 

1(0  H, 

36  10 

48)  48 

ADDITION  OF  FRACTIONS 

26.  Fractions  cannot  be  added  unless  they  have  a  common 
denominator.  We  cannot  add  1  to  -g-  as  they  now  stand,  since 
the  denominators  represent  parts  of  different  sizes.  Fourths 
cannot  be  added  to  eighths. 

Suppose  that  we  divide  an  apple  into  4  equal  parts,  and 
then  divide  2  of  these  parts  into  two  equal  parts.  It  is 
evident  that  we  shall  have  2  one-fourths  and  4  one-eighths. 
Now,  if  we  add  these  parts,  the  result  is  2  +  4  =  6  some¬ 
thing.  But  what  is  the  something?  It  is  not  fourths,  for 
six  fourths  are  li,  and  we  had  only  1  apple  to  begin  with; 
neither  is  it  eighths,  for  six  eighths  are  f,  which  is  less  than 
1  apple.  By  reducing  the  quarters  to  eighths,  we  have 
<-  =  I,  and  adding  the  other  4  eighths,  4  +  4  =  8  eighths. 


10 


ARITHMETIC 


§2 


This  result  is  correct,  since  1=1.  Or  we  can,  in  this  case, 
reduce  the  eighths  to  quarters.  Thus,  i  =  f;  whence,  adding, 
2  +  2  =  4  quarters,  a  correct  result,  since  1=1. 

Before  adding,  fractions  should  be  reduced  to  a  common 
denominator,  preferably  the  least  common  denominator. 

Example  1. — Find  the  sum  of  1,  f ,  and  f. 

Solution. — The  least  common  denominator,  or  the  least  number 
that  will  contain  all  the  denominators,  is  8. 

14  3  _  6  5  _  5 

2  8’  4  8’  aUd  8  8 


Explanation. — As  the  denominator  tells,  or  indicates, 
the  names  of  the  parts,  the  numerators  only  are  added,  to 
obtain  the  total  number  of  parts  indicated  by  the  denominator. 
Thus,  4  one-eighths  plus  6  one-eighths  plus  5  one-eighths  = 


4  6  5  4  +  6  +  515 

8  +  8  +  8  8  8 


li.  Ans. 


Example  2. — What  is  the  sum  of  121,  14f,  and  7 A? 

Solution.— The  least  common  denominator  in  this  case  is  16. 

12^  =  1241 
14|  =  141§ 

7A  =  T& 

sum  33  +  44  =  33  +  144  =  344+  Ans. 

The  sum  of  the  fractions  is  44  or  144,  which  added  to  the  sum  of  the 
whole  numbers  is  3444- 

Example  3. — What  is  the  sum  of  17,  13-&,  and  34? 

Solution. — The  least  common  denominator  is  32.  13^  =  13^, 

34  =  3A-  17 

13A 

9 

32 

3<ft 

sum  33f|  Ans. 

Rule. — I.  Reduce  the  given  fractions  to  fractions  having  the 
least  common  denominator  and  write  the  S7im  of  the  munerators 
over  the  cotmnon  denominator. 


II.  When  there  are  mixed  numbers  and  whole  numbers,  add 
the  fractions  first ,  and  if  their  stim  is  an  improper  fraction , 
reduce  it  to  a  mixed  number  and  add  the  whole  7iumber  with  the 
other  whole  mimbers. 


§2 


ARITHMETIC 


11 


EXAMPLES  FOR  PRACTICE 


Find  the  sum  of  the  following,  expressing  the  answers  in  their  lowest 
terms: 


)  A*  4i  and  3*2 • 

{b)  it,  -1%,  and  A- 

(c)  It,  M,  and  ft- 

( d )  10,  10i,  t,  and  |§. 

(e)  xt,  f.  and  ft. 

(f)  X)  ■§■)  and  A* 


Ans.  < 


(«)  To 

(b)  m 

(c)  3ff 

(d)  21| 

(e)  3| 

(/•)  Hi 


Note. — It  is  often  a  great  aid,  when  adding  fractions,  to  first  reduce  each  fraction 
to  its  lowest  terms. 


SUBTRACTION  OF  FRACTIONS 

27.  Fractions  must  first  have  a  common  denominator  before 
they  can  be  subtracted.  This  can  be  shown  in  the  same  man¬ 
ner  as  in  addition  of  fractions,'  Art.  26. 

Example  1. — Find  the  difference  between  to  and  f. 

Solution. — The  least  common  denominator  is  16. 

3  _  6  9  6  _  9  —  6  _  3 

8  ~  16’  16  16  “  16  “  16'  AnS' 

Example  2. — What  is  the  difference  between  6H  and  3yf? 
Solution. — The  least  common  denominator  of  the  fractions  is  36. 

fiii  _  ft33.  ol_3  _  026. 

Ol2  —  O30,  o  18  —  036 

minuend  6ff 
subtrahend  3ff 
difference  3^  Ans. 

Example  3. — From  12  take  9f. 

Solution. —  minuend  12  =  Ilf 

subtrahend  9 f  =  9f 
difference  2|  =  2|.  Ans. 

Explanation. — As  there  is  no  fraction  in  the  minuend 
from  which  to  take  the  fraction  in  the  subtrahend,  borrow  1, 
or  f,  from  12.  i  from  f  =  |.  Since  1  was  borrowed  from 
the  12,  11  is  left.  9  from  11  =  2.  2  -f  i  =  2gr.  Ans. 

Example  4. — From  9|  take  4A. 

Solution. — The  common  denominator  of  the  fractions  is  16.  9|=  9i%. 


minuend 

0-4- 
^1  0 

_  Q  20 

—  o16 

subtrahend 

4-L. 
^1  0 

—  4— 

—  ’10 

difference 

4Ui 

^10 

-  4|f.  Ans. 

12 


ARITHMETIC 


2 


Explanation. — As  the  fraction  in  the  subtrahend  is  greater 
than  the  fraction  in  the  minuend,  it  cannot  be  subtracted; 
therefore.,  borrow  1,  or  It,  from  the  9  in  the  minuend  and  add 
it  to  the  ~h\  its  +  It  =  ft.  tV  from  ft  =  It.  Since  1  was 
borrowed  from  9,  8  remains;  4  from  8  =  4;  4  +11  =  4|t. 

Rale. — I.  Reduce  the  fractions  to  fractions  having  a  com¬ 
mon  denominator.  Subtract  one  numerator  from  the  other  and 
write  the  remainder  over  the  common  denominator. 

II.  When  there  are  mixed  numbers ,  subtract  the  fractions 
and  whole  numbers  separately  and  place  the  remainders  side 
by  side. 

III.  When  the  fraction  in  the  subtrahend  is  greater  than  the 
fraction  in  the  minuend,  borrow  1  from  the  whole  ?i7imber  in  the 
minuend  and  add  it  to  the  fraction  in  the  minuend,  from  which 
subtract  the  fraction  in  the  subtrahend. 

IV.  When  the  minuend  is  a  whole  number,  borrow  1; 
reduce  it  to  a  fraction  whose  denominator  is  the  same  as  the 
denominator  of  the  fraction  in  the  subtrahend ,  and  place  it 
over  that  fraction  for  s7ibtraction. 


EXAMPLES  FOR  PRACTICE 


Solve  the  following  expressions: 

{a)  *-tf. 

(b)  ff-t. 

(C)  ~  it- 

(d)  16 1  -  8. 

{e)  20  -  lOf. 

(/)  18f-7*. 

(g)  17*  -  12f. 

(h)  349f  -  93*. 


Ans. 


(«)  i 

(b)  fV 

(c)  Mo 

(d)  8* 

(e)  9? 

(/)  HI 
(g)  4* 
(//)  2551 


1.  Which  is  the  larger,  1  or  H ?  Ans.  Equal 

2.  In  a  certain  mill  city,  1  of  the  looms  are  dobbies,  1  are  jac¬ 

quards,  and  |  are  box  looms,  the  remaining  are  all  plain  looms;  what 
fractional  part  of  the  whole  are  the  plain  looms?  Ans.  1 


§2 


ARITHMETIC 


13 


MULTIPLICATION  OF  FRACTIONS 

28.  When  multiplying  fractions,  it  is  not  necessary  to  reduce 
them  to  fractions  having  a  common  denominator. 


29.  Multiplying  the  numerator  or  dividing  the  denominator 
multiplies  the  fraction. 


Example. — Multiply  f  by  4. 


Solution. — 


Or, 


3 x  4  =  3X4 
8  X  8 

3  X  4  =  _ - _ 

8  X  8  -T  4 


-g-  =  H-  Ans. 
|  =  H.  Ans. 


30.  When  wishing  to  multiply  a  whole  number  by  a 
fraction,  the  method  is  the  same,  since  f  X  4  is  exactly  the 
same  in  value  as  4  X  i 


31.  The  word  of  when  used  in  connection  with  fractions 
means  multiplied  by  and,  consequently,  is  the  same  as  the 
multiplication  sign. 

Example  1. — What  is  §  of  40? 

c  5  ...  5  X  40  200  _o2 

Solution. —  ^of  40  =  — ^ =  224.  Ans. 

Example  2. — What  is  jv  of  f  ? 

e  11  ,  5  11  5  55  . 

Solution.-  12  of  6  =  12  X  6  =  72'  Ans' 

32.  Cancelation  may  often  be  used  to  advantage  in  the 
multiplication  of  fractions. 

Example  1. — What  is  f  of  50? 

25 

_  3  X  50  75  0_, 

Solution. —  — ^ —  371.  Ans. 

2 

Example  2. — What  is  the  product  of  f  and  |? 

_  3  5  15  5  A 

Solution. —  -  X  «  =  ht  =  5-  Ans. 

4  6  24  8 

Or,  by  cancelation, 

’3  5  _5_  =  5 
4  0  4X2  8‘ 

2 


Ans. 


14 


ARITHMETIC 


§2 


Example  3. — What  is  the  product  of  1,  -A,  and  f  ? 


Solution. — 


1111 
3  x  1  x  $  x  1  _  l 

4X3X12X3  2x3 

12  3  1 


1 

6' 


Ans. 


Example  4. — Multiply  41  by  3f. 

Solution.—  4\  =  f;  3f  =  ¥ 

?  x  ?7  243 

2  X  g  16  to  i  g . 


Ans. 


Example  5. — Multiply  21  by  8. 

Solution. —  2f  21 

8  or  8 

22  16  +  *£■  =  16  +  6  =  22.  Ans. 


Rule.— I.  Divide  the  product  of  the  numerators  by  the  prod¬ 
uct  of  the  denominators .  All  factors  common  to  both  numera¬ 
tors  and  denominators  should  first  be  canceled. 


II.  To  multiply  a  mixed  number  by  a  mixed  number  reduce 
them  to  improper  fractions  and  Proceed  as  with  the  multiplica¬ 
tion  of  fractions. 

III.  To  multiply  a  mixed  number  by  a  whole  number ,  first 
multiply  the  whole  numbers  together  and  to  this  Product  add  the 
product  of  the  fractional  part  and  the  multiplier.  Should  this 
last  result  be  an  improper  fraction ,  first  reduce  it  to  a  mixed 
number  and  then  add  the  result  to  the  product  of  the  whole 
numbers. 


EXAMPLES  FOB  PRACTICE 

Solve  the  following  expressions,  giving  the  answers  in  lowest  terms: 


(a) 

3.  w  1 

8  A  2  • 

[(*) 

3 

TS 

(b) 

0A2A4A8A  10* 

(b) 

1 1 1 

Ire 

ic) 

84  X  f. 

(c) 

63 

( d ) 

1  X  100. 

(d) 

83* 

(e) 

151X20.  Ans-' 

(e) 

315 

(f) 

14  X  10*. 

( f ) 

147 

U) 

41  X  9*. 

Or) 

451 

(h) 

251  X  HI. 

lw 

9QQ XL 

cLXjV  3  2 

2 


ARITHMETIC 


15 


DIVISION  OF  FRACTIONS 

33.  Division  of  fractions  is  the  exact  reverse  of  multiplica¬ 
tion  of  fractions.  In  division  of  fractions ,  it  is  not  necessary  to 
reduce  the  fractio?is  to  those  having  a  common  denominator. 

34.  To  divide  a  fraction  by  a  whole  number: 


Rule. — Divide  the  numerator  or  multiply  the  denominator 
by  the  whole  number. 


Example.— Divide  f  by  2. 

Solution. — Dividing  the  numerator  by  the  whole  number  gives 


4  4-f-2  _  2 

8  ‘  ~  8  “8 


1 

4' 


Ans. 


Multiplying  the  denominator  by  the  whole  number  gives 

4 


-  -h  2  = 

8  8X2 


4  1  A 

=  16  =  4‘  AnS- 


35.  To  divide  a  whole  number  or  a  fraction  by  a 
fraction: 


Rule. — Invert  the  divisor  and  proceed  as  in  multiplication. 

Example  1. — Divide  16  by  f. 

4  fi  QA 

Solution.—  16  4-  ^  =  16  X  t  =  24.  Ans. 

6  4  4 


To  invert  a  fraction  is  to  turn  it  upside  down;  that  is, 
make  the  numerator  and  denominator  change  places. 

Invert  1  and  it  becomes 


Example  2. — Divide  la  by  t%. 

Solution. — 1.  The  fraction  la  is  contained  in  ^  3  times,  for  the 
denominators  are  the  same,  and  one  numerator  is  contained  in  the 
other  3  times. 

2.  If  we  now  invert  the  divisor,  i%,  and  multiply,  the  solution  is 


9  16  =  0>O0  . 

16  X  3  10  X  0  6'  A 

This  brings  the  same  quotient  as  in  the  first  case. 


Example  3. — Divide  f  by  j. 

Solution. — We  cannot  divide  f  byi,  as  in  the  first  case  of  example 
2,  for  the  denominators  are  not  the  same;  therefore,  we  must  solve  as 
in  the  second  case. 

3^1=3  4=3X4  3 

8  ’  4  8*1 


8X1 

2 


Ans. 


16 


ARITHMETIC 


§2 


Example  4. — Divide  5  by  fi. 


Solution. — 


1  0 
1  6 


inverted  becomes 
8 

_  i6  9  x  n 

5  x  lo  =  ~vr 

t 


1  6 
1  O  • 


=  8. 


Ans. 


36.  Whenever  a  mixed  number  has  to  be  dealt  with  in 
the  division  of  fractions,  the  best  method  is  to  change  the 
mixed  number  to  an  improper  fraction  and  then  proceed 
according  to  the  foregoing  rules. 

Example  1. — How  many  times  is  3f  contained  in  7re? 


Solution.- 


0  3  1  5 . 

“  04  —  4  , 

119  4  _ 

16  15 


7A  = 


119 
1  6  • 


119  X  4 

16  X  15 
4 


inverted  equals  A- 
119  e 

60  8 


Ans. 


Example  2. — Divide  5|  by  2|. 
Solution.—  5|  =  V;  =  f . 


Then, 


26^_5  _  26  2 

5  '  2  ~  5*5 


52 

25 


—  9-A 

—  ^26* 


Ans. 


EXAMPLES  FOR  PRACTICE 


Divide: 

(a)  15  by  6f . 

(£)  30  by!. 

(c)  172  by  |. 

(d)  by  lAu 

(e)  ^byl4f. 
(/)  W  by  17| 


Ans.  • 


(а)  2| 

(б)  40 

(c)  215 

(d)  m 

(e)  Hi 


1(0 


37.  As  explained  in  Art.  10,  when  a  fraction  contains  a 

fractional  number  for  its  numerator  or  denominator,  or  both, 

i  i  4 

it  is  known  as  a  complex  fraction.  Thus.  A  and  -  are  all 

8  s  "3 

complex  fractions. 

38.  It  will  be  noticed  in  the  above  illustrations  of  com¬ 
plex  fractions  that  a  heavier  line  is  used  to  separate  parts  of 
the  fractions.  This  heavy  line  always  separates  the  numer¬ 
ator  and  denominator  of  complex  fractions.  Thus,  in  the 
first  case,  f  is  the  numerator  and  8  the  denominator.  In 


§2 


ARITHMETIC 


17 


the  second  fraction,  £  is  the  numerator  and  i  the  denomi¬ 
nator.  In  the  third  fraction,  4  is  the  numerator  and  f,  the 
denominator. 

A  line  drawn  between  two  numbers  in  this  manner  indi¬ 
cates  that  the  one  above  the  line  is  to  be  divided  by  the 
one  below  the  line.  Consequently,  in  order  to  reduce  a 
complex  fraction  to  a  common  one,  divide  the  numerator 
by  the  denominator. 

3. 

Example. — Reduce  £  to  a  common  fraction. 

O 

c  i  3  _  3  1  3  . 

Solution.—  g=|  +  8  =  |Xg=g2.  Ans. 


39.  Whenever  an  expression  like  one  of  the  three  fol¬ 
lowing  is  obtained,  it  may  always  be  simplified  by  trans¬ 
posing  the  denominator  from  above  to  below  the  line,  or 
from  below  to  above,  as  the  case  may  be,  taking  care,  how¬ 
ever,  to  indicate  that  the  denominator  when  so  transferred  is 
a  multiplier. 

3  1 

1.  -  — -  =  —  =  — :  for,  regarding  the  fraction  above 

9  9  X  4  36  12 

the  heavy  line  as  the  numerator  of  a  fraction  whose  denom¬ 
inator  is  9,  -  X  ^  as  before. 

9X4  9X4’ 

2.  |  =  ^  —  12.  The  proof  is  the  same  as  in  the 

4  O 


first  case. 


O  I  =  5X4 
'I  3X9 


20 

27’ 


For,  regarding  f  as  the  numerator 


5  x  9 

of  a  fraction  whose  denominator  is  1,  4 

t  X  9 


3X9’ 


and 


5  X4  5X4  20  , 

-  =  as  above. 

3  X  9  x  4  3  X  9  27’ 


This  principle  may  be  used  to  great  advantage  in  cases  like 


X  310  X  n  X  72 
40  X  4i  X  5 i 


Reducing  the  mixed  numbers  to  fractions, 


18 


ARITHMETIC 


§2 


.  ,  i  X  310  X  fi  X  72  ,T  ,  f  . 

the  expression  becomes  . — .  Now,  transferring 

40xfx¥ 

the  denominators  of  the  fractions  and  canceling, 

3 

X0  3  0  3 

1  X  310  X  27  X  72  X  2  X  6  =  1  X  340  X  X  U  X  %  X  0 

40  X  9  X  31  X  4  X  12  40  X  0  X  34  X  4  X  42 

4  2 

=  —  =  13i  ^ 


Greater  exactness  in  results  can  usually  be  obtained  by 
using  this  principle  than  by  reducing  the  fractions  to  deci¬ 
mals.  The  principle,  however,  should  not  be  employed  if  a 
sign  of  addition  or  subtraction  occurs  either  above  or  below 
the  dividing  line. 


DECIMALS 

40.  Decimals  are  tenth  fractions;  that  is,  the  parts  of  a 
unit  are  expressed  on  the  scale  of  ten,  as  tenths,  hundredths , 
thousandths ,  etc. 

41.  The  denominator,  which  is  always  ten  or  a  multiple 
of  ten,  as  10,  100,  1,000,  etc.,  is  not  expressed,  as  it  would  be 
in  common  fractions,  by  writing  it  under  the  numerator  with 
a  line  between  them,  as  mf,  tot,  toVo,  but  is  expressed  by 
placing  a  period  (.),  which  is  called  a  decimal  point,  to 
the  left  of  the  figures  of  the  numerator,  so  as  to  indicate 
that  the  number  on  the  right  is  the  numerator  of  a  fraction 
whose  denominator  is  10,  100,  1,000,  etc. 

42.  The  reading  of  a  decimal  number  depends  on  the 
number  of  decimal  places  in  it,  or  the  number  of  figures 
to  the  right  of  the  decimal  point. 

One  decimal  place  expresses  tenths 
Two  decimal  places  express  hundredths 
Three  decimal  places  express  thousandths 
Four  decimal  places  express  ten-thousandths 
Five  decimal  places  express  hundred-thousandths 
Six  decimal  places  express  millionths 


2 


ARITHMETIC 


19 


Thus: 

.3 

3 

1  0 

=  3  tenths 

.03 

3 

10  0 

=  3  hundredths 

.003 

s 

10  0  0 

=  3  thousandths 

.0003  = 

3 

1  0  0  0  0 

=  3  ten-thousandths 

.00003  = 

3 

100000 

=  3  hundred-thousandths 

.000003  = 

3 

1000000 

=  3  millionths 

We  see  in  the  above  that  the  number  of  decimal  places  in  a 
decimal  equals  the  number  of  ciphers  to  the  right  of  the  figure  1  in 
the  denominator  of  its  equivalent  fraction.  This  fact  kept  in  mind 
will  be  of  much  assistance  in  reading  and  writing  decimals. 

43.  Whatever  may  be  written  to  the  left  of  a  decimal 
point  is  a  whole  number.  The  decimal  point  merely  separates 
the  fraction  on  the  right  from  the  whole  number  on  the  left. 

44.  When  a  whole  number  and  decimal  are  written 
together,  the  expression  is  a  mixed  number.  Thus,  8.12  and 
17.25  are  mixed  numbers. 


45.  The  relation  of  decimals  and  whole  numbers  to  each 
other  is  clearly  shown  by  the  following  table: 


m 

G 


m 

G 


o 

•  T-( 

m 

G 

C/3 

•  i-H 

tn 

O 

C 

a 

£ 

o 

X 

4-1 

a 

C/3 

•4-4 

•  r-i 

>4-4 

3 

O 

•  T—C 

O 

o 

m 

a 

in 

m 

XI 
-*— > 

<D 

G 

CD 

MX 

J-4 

O 

o 

t-4 

o 

T3 

G 

C/3 

r~* 

•  T- < 

T3 

G 

C/3 

G 

X 

53 

+-» 

•  T—4 

a 

G 

X 

53 

+-» 

9 

8 

7 

6 

5 

m 

T3 

G 

ctf 

in 

G 

O 

XI 


m 

CD 

i— i 

TG 

G 

G 

X 


m 

G 

<D 


m 

4-j 

•  T—< 

G 

G 


O 

a 

in 

G  X 


a 

<d 


G 

CD 


in 

X 

-i-i 

HG 

<D 

J-4 

TG 

G 

G 

X 


in 

X 

-i-> 

TG 

G 

aJ 

in 

G 

O 

X 


m 

X 

-i-> 

ra 

G 

rt 

m 

G 

O 

X 

4-1 

I 

G 

CD 


CO 

X 

4J 

TG 

G 

a] 

in 

G 

O 

X 

4-1 

I 

"O 

CD 

U 

TG 

G 

G 

X 


m 

X 

4-> 

G 

O 


2  3  4  5 


a 

6  7 


in 

X 

4-1 

G 

O 


G 

<D 


CO 

X 

4-1 

G 

O 


TG 

CD 

1-4 

TG 

G 

G 

X 


8  9 


The  number  represented  by  the  figures  to  the  left  of  the 
decimal  point  is  a  whole  number;  that  to  the  right  is  a  decimal. 

46.  In  both  the  decimals  and  whole  numbers,  the  units 
place  is  made  the  starting  point  of  notation  and  numeration. 
Both  whole  numbers  and  decimals  decrease  on  the  scale  of 
ten  to  the  right,  and  both  increase  on  the  scale  of  ten  to  the 
left.  The  first  figure  to  the  left  of  units  is  tens ,  and  the  first 


20 


ARITHMETIC 


§2 


figure  to  the  right  of  units  is  tenths.  The  second  figure  to 
the  left  of  units  is  hundreds ,  and  the  second  figure  to  the 
right  is  hundredths .  The  third  figure  to  the  left  is  thousands, 
and  the  third  to  the  right  is  thousandths,  and  so  on;  the 
whole  numbers  on  the  left  and  the  decimals  on  the  right. 
The  figures  equally  distant  from  units  place  correspond  in 
name,  the  decimals  having  the  ending  ths,  to  distinguish 
them  from  whole  numbers.  The  following  is  the  numeration 
of  the  number  in  the  above  table:  nine  hundred  eighty-seven 
million,  six  hundred  fifty-four  thousand,  three  hundred  twenty- 
one  and  twenty-three  million,  four  hundred  fifty-six  thousand, 
seven  hundred  eighty-nine  hundred-millionths. 

The  decimals  increase  to  the  left,  on  the  scale  of  ten,  the 
same  as  whole  numbers;  for,  if  you  begin  at  the  4  'in  thou¬ 
sandths  place  in  the  above  table,  the  next  figure  to  the  left 
is  hundredths,  which  is  ten  times  as  great,  and  the  next 
tenths,  or  ten  times  the  hundredths,  and  so  on  through  both 
decimals  and  whole  numbers. 


47.  Annexing  or  taking  away  a  cipher  at  the  right  of  a 
decimal  does  not  affect  its  value. 

.5  is  .50  is  but  —  =  therefore,  .5  =  .50 

10  100  10  100 

48.  Inserting  a  cipher  between  a  decimal  and  the  decimal 
point ,  divides  the  decimal  by  10. 


.5  =  A;  A  +  10  =  A- 

10  10  100 


=  .05 


49.  Taking  away  a  cipher  from  the  left  of  a  decimal  multi¬ 
plies  the  decimal  by  10. 

.05  =  JL:  ~  X  10  =  A  =  .5 


100’  100 


50.  In  some  cases,  it  is  convenient  to  express  a  mixed 
decimal  fraction  in  the  form  of  a  common  (improper)  fraction. 
To  do  so  it  is  only  necessary  to  write  the  entire  number, 
omitting  the  decimal  point,  as  the  numerator  of  the  frac¬ 
tion,  and  the  denominator  of  the  decimal  part  as  the  denom¬ 
inator  of  the  fraction.  Thus,  127.483  =  loo3’,  for  127.483 

m  4  8  3  127000  +  483  1  2  7  4  8  3 

1  00  0  —  1  o  0  0  =  1  000  . 


2 


ARITHMETIC 


21 


ADDITION  OF  DECIMALS 

51.  Addition  of  decimals  is  similar  in  all  respects  to 
addition  of  whole  numbers — units  are  placed  under  units,  tens 
under  tens,  etc.;  this,  of  course,  brings  the  decimal  points  in 
line,  directly  under  one  another.  Hence,  in  placing  the  num¬ 
bers  to  be  added,  it  is  only  necessary  to  take  care  that  the 
decimal  points  are  in  line.  In  adding  whole  numbers,  the 
right-hand  figures  are  always  in  line;  but  in  adding  decimals, 
the  right-hand  figures  will  not  be  in  line  unless  each  decimal 
contains  the  same  number  of  figures. 


who'.e  numbers 

decimals 

mired  numbers 

3  4  2 

.3  4  2 

3  4  2.0  3  2 

4  2  3  4 

.4  2  3  4 

4  2  3  4.5 

2  6 

.2  6 

2  6.6  7  8  2 

3 

.0  3 

3.0  6 

sum  4  6  0  5  Ans. 

sum  1.0  5  5  4  Ans. 

sum  4  6  0  6.2  7  0  2 

Example. — What  is  the  sum  of  242,  .36,  118.725,  1.005,  6,  and  100.1? 

Solution. —  2  4  2. 

.3  6 

1  1  8.7  2  5 
1.0  0  5 
6. 

1  0  0.1 

sum  4  6  8.1  9  0  Ans. 

Rule. — Place  the  numbers  to  be  added  so  that  the  decimal 
Points  will  be  directly  under  one  another.  Add  as  in  whole 
numbers ,  and  place  the  decimal  point  in  the  sum  directly  under 
the  decimal  points  above. 


EXAMPLES  FOR  PRACTICE 
Find  the  sum  of: 

(a)  999.999,  66.66,  4.4,  and  .43.  Ans.  1,071.489 

(b)  277.36,  306.005,  24.05,  and  940.403.  Ans.  1,547.818 

(c)  146.04,  .0005,  500,  and  .711.  Ans.  646.7515 

(d)  Nine  tenths;  forty-four  hundredths;  ninety-six  and  eight  tenths; 

and  four  hundred  eleven  and  sixty-nine  hundredths.  Ans.  509.83 


22 


ARITHMETIC 


§2 


SUBTRACTION  OF  DECIMALS 

52.  As  in  subtraction  of  whole  numbers,  units  are  placed 
under  units,  tens  under  tens,  etc.,  bringing  the  decimal  points 
under  each  other,  as  in  addition  of  decimals. 

Example  1. — Subtract  .132  from  .3063. 

Solution. —  minuend  .3  0  6  3 

subtrahend  .13  2 

difference  .1  7  4  3  Ans. 

Example  2. — What  is  the  difference  between  7.895  and  .725? 

Solution. —  minuend  7.8  9  5 

subtrahend  .7  2  5 

difference  7.1  7  0  or  7.17  Ans. 

Example  3. — Subtract  .625  from  11. 

Solution. —  minuend  1  1.0  0  0 

subtrahend  .6  2  5 

difference  1  0.3  7  5  Ans. 

Rule. — Place  the  subtrahend  under  the  minuend,  so  that  the 
decimal  Points  will  be  directly  under  each  other.  Subtract,  as  in 
whole  numbers,  and  place  the  decimal  point  in  the  remainder, 
directly  under  the  decimal  points  above. 

When  the  figures  in  the  decimal  part  of  the  subtrahend 
extend  beyond  those  in  the  minuend,  place  ciphers  in  the  minuend 
above  them  and  subtract  as  before. 


EXAMPLES  FOR  PRACTICE 


From: 

(a) 

407.385  take  235.0004. 

(b) 

22.718  take  1.7042. 

(c) 

1,368.17  take  13.6817. 

id) 

70.00017  take  7.000017. 

(e) 

630.630  take  .6304. 

(0 

421.73  take  217.162. 

(g) 

1.000014  take  .00001. 

( h ) 

.783652  take  .542314. 

\{a) 

172.3846 

(b) 

21.0138 

(c) 

1,354.4883 

(d) 

63.000153 

(e) 

629.9996 

(0 

204.568 

(g) 

1.000004 

Iw 

.241338 

2 


ARITHMETIC 


23 


MULTIPLICATION  OF  DECIMALS 

53.  In  multiplication  of  decimals,  it  is  not  necessary  to 
have  the  decimal  points  directly  below  one  another.  Multi¬ 
plication  of  decimals  is  performed  exactly  the  same  as  the 
multiplication  of  whole  numbers,  with  the  exception  of  pla¬ 
cing  the  decimal  point  in  the  answer.  After  multiplying, 
count  the  number  of  decimal  places  in  both  multiplicand  and 
multiplier,  and  point  off  the  same  number  of  decimal  places  in 
the  product. 

Example. — Multiply  3.38  by  .5. 

Solution. —  multiplicand  3.3  8 

multiplier  .5 

product  1.6  9  0  Ans. 

Explanation. — In  this  example  there  are  2  decimal  places 
in  the  multiplicand  and  1  in  the  multiplier;  therefore,  2  +  1 
=  3  decimal  places  are  pointed  off  in  the  product. 

54.  It  sometimes  happens  that  the  number  of  figures  in 
the  answer  will  not  equal  the  sum  of  the  decimal  places  in 
the  numbers  multiplied,  in  which  case  ciphers  must  be  placed 
at  the  left  of  the  answer. 

Example. — Multiply  .346  by  .005. 

Solution. —  multiplicand  .3  4  6 

multiplier  .0  0  5 

product  .0  0  1  7  3  0  Ans. 

Explanation. — Multiplying  346  by  5  gives  1,730,  but 
there  are  3  decimal  places  in  the  multiplicand  and  3  in  the 
multiplier;  therefore,  there  must  be  3  +  3  =  6  decimal  places 
in  the  answer.  The  number  1,730  contains  only  4  figures; 
therefore,  it  is  necessary  to  prefix  2  ciphers  to  1,730  in  order 
to  give  the  six  decimal  places.  The  result  is  .001730,  or 
.00173.  Ans. 

55.  The  reason  for  adding  the  number  of  decimal  places 
in  the  multiplier  and  multiplicand  to  find  out  how  many 
decimal  places  to  point  off  in  the  product  will  be  readily 
apparent  from  the  following: 


24 


ARITHMETIC 


§2 


Consider  the  numbers  in  the  last  example,  and  express  the 
multiplier  and  multiplicand  in  the  form  of  common  fractions, 
obtaining,  respectively,  and  two.  The  product  of  these 
two  fractions  is 


346  5  =  846  X  5 

1000  X  1000  1000  X  1000 


1730 

1000000 


.001730  =  .00173 


It  will  be  noticed  that  the  denominator  of  the  fraction 
expressing  the  product  has  as  many  ciphers  in  it  as  is 
expressed  by  the  sum  of  the  ciphers  in  the  multiplier  and 
multiplicand;  as  this  is  always  the  case,  and  as  the  number 
of  decimal  places  in  any  decimal  is  equal  to  the  number  of 
ciphers  in  its  denominator  when  expressed  as  a  common 
fraction,  it  is  evident  that  the  number  of  decimal  places  in 
the  product  is  equal  to  the  sum  of  the  decimal  places  in 
the  multiplier  and  multiplicand. 


Hale. — Multiply  the  decimals  as  in  whole  numbers  and  point 
off  as  many  decimal  places  in  the  product  as  there  are  decimal 
places  in  both  the  multiplier  and  multiplicand ,  prefixing  ciphers 
if  necessary. 


EXAMPLES  FOR  PRACTICE 


Find  the  product  of: 

(a)  20.05  X  4. 

(< b )  18.34X4.5 

(c)  342  X. 007. 

(d)  987  X  .25. 
(<?)  36  X  .36. 
(/)  550  X. 55. 

(g)  i.n  X. 11. 

(h)  .364  X  .003 


Ans. 


'{a)  80.2 
{b)  82.53 

(c)  2.394 

(d)  246.75 

(e)  12.96 
(/)  302.5 
ig)  -1221 

.(/z)  .001092 


DIVISION  OF  DECIMALS 

56.  In  division  of  decimals  the  number  of  decimal  places 
in  the  dividend  must  equal  or  exceed  (or  be  ?nade  to  equal  or 
exceed  by  annexing  ciphers )  the  number  of  decimal  places  in 
the  divisor.  Divide  exactly  as  in  whole  numbers.  Subtract 
the  number  of  decimal  places  in  the  divisor  from  the  number  of 


2 


ARITHMETIC 


25 


decimal  Places  in  the  dividend ,  and  point  off  as  many  decimal 
places  in  the  quotient  as  there  are  units  in  the  remainder 
thus  found. 

In  the  following  examples,  which  illustrate  every  case  that 
can  arise  in  division  of  decimals,  all  the  results  are  proved 
by  expressing  the  decimals  in  the  form  of  common  fractions 
before  dividing: 


Example  1. — Divide  .625  by  25. 
Solution.— 


divisor  dividend  Quotient 

2  5  )  .6  2  5  (  .0  2  5  Ans. 
5  0 


remainder 

Proof. — 

.625  ^  25  = 


12  5 
1  2 

0 


625 


1000 


-  25  = 


25 

m 


i 


iooo  x  n 


25 

1000 


=  .025 


In  this  example  there  are  no  decimal  places  in  the  divisor, 
and  3  decimal  places  in  the  dividend;  therefore,  there  are 
3  minus  0,  or  3,  decimal  places  in  the  quotient.  One  cipher 
must  be  prefixed  to  the  25,  to  make  the  three  decimal  places. 


Example  2. — Divide  6.035  by  .05. 
Solution. — 


divisor  dividend  quotient 

.0  5  )  6.0  3  5  (  1  2  0.7  Ans. 
5 


Proof. — 
6.035  --  .05  = 


l  0 
10 

3  5 
3_5 

remainder  0 

6035  .  5 


1000  100 


1207 

$m  XM- 1207 


tm 

10 


10 


=  120.7 


In  this  example,  divide  by  5,  as  if  the  cipher  were  not 
before  it.  There  is  one  more  decimal  place  in  the  dividend 
than  in  the  divisor;  therefore,  one  decimal  place  is  pointed 
off  in  the  quotient. 


26 


ARITHMETIC 


§2 


Example  3. — Divide  .125  by  .005. 


divisor  dividend  quotient 

Solution. —  .0  05)  .1  25(25  Ans. 

10 
2  5 
25 

remainder  0 


Proof. — 


.125  -r  .005  = 


25 

125  .  5  =  m  y 

1000  '  1000  X000 


1000 

$ 


25 


In  this  example  there  are  the  same  number  of  decimal 
places  in  the  dividend  as  in  the  divisor.  In  this  case  the 
quotient  has  no  decimal  places  and  is  a  whole  number. 


Example  4. — Divide  326  by  .25. 


divisor  dividend  quotient 

Solution. —  .2  5  )  3  2  6.0  0  (  1  3  04  Ans. 

25 
7  6 
7_5 
10  0 
10  0 

remainder  0 


Proof. — 

326  4-  .25  =  326  —  =  326  X  — 

100  25 


32600 

25 


=  1304 


In  this  problem  two  ciphers  were  annexed  to  the  dividend, 
to  make  the  number  of  decimal  places  equal  to  the  number 
in  the  divisor.  The  quotient  is  a  whole  number. 


Example  5.— Divide  .0025  by  1.25. 


Solution. — 


Proof. — 


divisor  dividend  quotient 

1.2  5  )  .0  0  2  5  0  (  .0  0  2  Ans. 
2  5  0 

remainder  0 


.0025  4-  1.25  = 


25  .  125 

10000  '  100 


1  =  _2_ 
500  1000 


.002 


J_YI 

10000  A  m 

100  5 


Explanation. — In  this  example  we  are  to  divide  .0025 
by  1.25.  Consider  the  .dividend  as  a  whole  number,  or  25 
(disregarding  the  two  ciphers  at  its  left,  for  the  present); 


2 


ARITHMETIC 


27 


also,  consider  the  divisor  as  a  whole  number  or  125.  It  is 
evident  that  the  dividend  25  will  not  contain  the  divisor  125; 
we  must,  therefore,  annex  one  cipher  to  the  25,  thus  making 
the  dividend  250.  125  is  contained  twice  in  250,  so  we  place 

the  figure  2  in  the  quotient.  In  pointing  off  the  decimal 
places  in  the  quotient,  it  must  be  remembered  that  there 
were  only  4  decimal  places  in  the  dividend;  but  one  cipher 
was  annexed,  thereby  making  4  +  1,  or  5,  decimal  places. 
Since  there  are  5  decimal  places  in  the  dividend  and  2  decimal 
places  in  the  divisor,  we  must  point  off  5  —  2,  or  3,  decimal 
places  in  the  quotient.  In  order  to  point  off  3  decimal  places, 
two  ciphers  must  be  prefixed  to  the  figure  2,  thereby  making 
.002  the  quotient.  It  is  not  necessary  to  consider  the  ciphers 
at  the  left  of  a  decimal  when  dividing,  except  when  deter¬ 
mining  the  position  of  the  decimal  point  in  the  quotient. 

Rule. — I.  Place  the  divisor  to  the  left  of  the  dividend ,  and 
proceed  as  in  division  of  whole  numbers;  in  the  quotient ,  point 
off  as  many  decimal  places  as  the  number  of  decimal  places  in  the 
dividend  exceed  those  in  the  divisor ,  prefixing  ciphers  to  the 
quotient ,  if  necessary. 

II.  If  in  dividing  one  number  by  another  there  be  a  remain¬ 
der ,  the  remainder  can  be  placed  over  the  divisor ,  as  a  fractional 
part  of  the  quotient ,  but  it  is  generally  better  to  annex  ciphers  to 
the  remainder ,  and  continue  dividing  until  there  are  3  or  4  deci¬ 
mal  places  in  the  quotient ,  and  then  if  there  still  be  a  remainder , 
terminate  the  quotient  by  the  plus  sign  (  +  ),  which  shows  that 
it  can  be  carried  further. 

57.  It  frequently  happens,  as  in  the  following  example, 
that  the  division  will  never  terminate.  In  such  cases,  decide 
to  how  many  decimal  places  the  division  is  to  be  carried, 
and  carry  the  work  one  place  further.  If  the  last  figure 
of  the  quotient  thus  obtained  is  5  or  a  greater  number, 
increase  the  preceding  figure  by  1,  and  write  after  it  the 
minus  sign  (  — ),  thus  indicating  that  the  quotient  is  not  quite 
as  large  as  indicated;  if  the  figure  thus  obtained  is  less 
than  5,  write  the  plus  sign  ( +- )  after  the  quotient,  thus  indi¬ 
cating  that  the  number  is  slightly  greater  than  as  indicated. 


28 


ARITHMETIC 


§2 


In  the  following  example,  had  it  been  desired  to  obtain  the 
correct  answer  to  four  decimal  places,  the  work  would  have 
been  carried  to  five  places,  obtaining  13.26666,  and  the 
answer  would  have  been  given  as  13.2667  —  .  This  remark 
applies  to  any  other  calculation  involving  decimals,  when  it 
is  desired  to  omit  some  of  the  figures  in  the  decimal.  Thus, 
if  it  is  desired  to  retain  three  decimal  places  in  the  num¬ 
ber  .2471253,  it  would  be  expressed  as  .247  +  ;  if  it  was 
desired  to  retain  five  decimal  places,  it  would  be  expressed 
as  .24713  —  .  Both  the  +  and  —  signs  are  frequently  omitted; 
they  are  seldom  used  except  in  exact  calculations,  when  it  is 
desired  to  call  particular  attention  to  the  fact  that  the  result 
obtained  is  not  quite  exact. 

Example. — What  is  the  quotient  of  199  divided  by  15? 

Solution. —  15)199(13-+  ^  =  13A  Ans. 

1  5 
4  9 
4  5 
4 

Or,  1  5  )  1  9  9.0  0  0  (  1  3.2  6  7  -  Ans. 

1  5 
4  9 
4  5 
4  0 
3  0 
10  0 
9  0 
1  00 
9  0 
1  0 


EXAMPLES  FOR  PRACTICE 

Divide: 


(«) 

101.6688  by  2.36. 

f(«) 

43.08 

(6) 

187.12264  by  123.107. 

(*) 

1.52 

(c) 

.08  by  .008. 

(c) 

10 

(d) 

.0003  by  3.75.  * 

(d) 

.00008 

(*) 

.0144  by  .024.  nS" 

(e) 

.6 

(f) 

.00375  by  1.25. 

(H 

.003 

Or) 

.004  by  400. 

Gr) 

.00001 

(h) 

.4  by  .008. 

1(A) 

50 

2 


ARITHMETIC 


29 


TO  REDUCE  A  FRACTION  TO  A  DECIMAL 


58.  The  following  examples  and  rule  show  how  a  value 
expressed  as  a  common  fraction  may  be  changed  to  a  decim'al: 


Example  1. — 
Solution. — 


f  equals  what  decimal? 


4  )  3.0  0 
.7  5 


or  |  =  .75. 


Ans. 


Example  2. — What  decimal  is  equivalent  to  f  ? 
Solution. —  8  )  7.0  0  0  (  .8  7  5 

6  4 


6  0 

5  6  or  1  =  .875.  Ans. 

4  0 
4  0 

0 

Rule. — Annex  ciphers  to  the  numerator  and  divide  by  the 
denominator.-  Point  off  as  many  decimal  places  in  the 
quotient  as  there  are  ciphers  annexed. 


EXAMPLES  FOR  PRACTICE 
Reduce  the  following  common  fractions  to  decimals: 

(a)  H.  fO)  -46875 

(b)  I- 

(c)  ih. 

(d)  fi. 

(e) . 

(0  I- 


Ans.' 


Or)  21o°o  • 
(h)  T5TT 


(6)  .875 
{c)  .65625 

(d)  .796875 

(e)  .16 
(/)  .625 
G r)  .05 
{h)  .004 


59.  To  reduce  inches  to  a  decimal  part  of  a  foot: 
Example. — What  decimal  part  of  a  foot  is  9  inches? 

Solution. — Since  there  are  12  in.  in  1  ft. ,  1  in.  is  of  a  foot,  and  9  in. 
is  9  X  A  —  &  ft.  This  reduced  to  a  decimal  by  the  above  rule,  shows 
what  decimal  part  of  a  foot  9  in.  is. 

1  2  )  9.0  0  (  .7  5  ft.  Ans. 


30  ARITHMETIC  §2 

Rule. — I.  To  reduce  inches  to  decimal  parts  of  a  foot,  divide 

the  number  of  inches  by  12. 

II.  Should  the  resulting  decimal  be  an  unending  07ie ,  and  it  is 
desired  to  terminate  the  division  at  some  point ,  say  the  fourth  deci- 
77i al  place ,  carry  the  division  07ie  place  further ,  and  if  the  fifth 
figure  is  5  or  greater,  hicrease  the  foiu'th  figure  by  1. 


EXAMPLES  FOR  PRACTICE 


Reduce  to  the  decimal  part  of  a  foot: 

(a)  3  inches. 

(b)  4^  inches. 

( c )  5  inches. 

( d )  6f  inches. 

(e)  11  inches. 


(a)  .25  ft. 


Ans.< 


(b)  .375  ft. 
(t)  .4167  ft. 
( d )  .5521  ft. 
(< e )  .9167  ft. 


TO  REDUCE  A  DECIMAL  TO  A  FRACTION 
60.  The  following  examples  and  rule  show  how  a  value 
expressed  as  a  decimal  may  be  changed  to  a  common  fraction: 


Example  1. — Reduce  .125  to  a  fraction. 


Solution.— 


.125 


125 

1,000 


Ans. 


Example' 2. — Reduce  .875  to  a  fraction. 

■  875  35  7 

Solution.-  .875  =  “  40  =  8'  Ans' 

Rule. — U7ider  the  figures  of  the  decimal,  place  1  with  as 
many  ciphers  at  its  right  as  there  are  deci77ial  places  in  the  dec¬ 
imal,  and  rediice  the  resulting  fractio7i  to  its  lowest  terms  by 
dividmg  both  mi77ierator  a7id  de7io7ninator  by  the  sa77ie  mimber. 


EXAMPLES  FOR  PRACTICE 


Reduce  the  following  to  common  fractions: 


(a)  .125. 

(b)  .625. 

(c)  .3125. 

(d)  .04. 

(e)  .06. 
(/)  .75. 
(g)  -15625 
{h)  .875. 


Ans. 


(a) 

(b) 

(c) 

(d) 

(e) 
(/) 
(g) 

[(h) 


x 

8 

5. 

8 
_B 
1  6 
1 

2T 

3 

60 

3. 

4 


2 


ARITHMETIC 


31 


61.  To  express  a  decimal  approximately  as  a  frac¬ 
tion  having  a  given  denominator: 

Example  1. — Express  .5827  in  64ths. 


Solution. - 


cotV7w*~  37.2928 
.5827  X  777  =  — 7.7 — ,  say 
o4 


64 

64 


37 

64 


Hence,  .5827  =  — ,  nearly.  Ans. 

o4 

Example  2. — Express  .3917  in  12ths. 

c  9m7v  12  4.7004  5 

Solution.—  .3917  X  ^  =  ~i2~>  sa^  12 

g 

Hence,  .3917  =  — ,  nearly.  Ans. 

1Z 


Rule. — Reduce  1  to  a  fractio?i  having  the  given  denominator. 
Multiply  the  given  decimal  by  the  fraction  so  obtained,  and  the 
result  will  be  the  fraction  required. 


Express: 

(a) 

(b) 

(c) 

(d) 

(e) 

(f) 


EXAMPLES  FOR  PRACTICE 


.625  in  8ths. 
.3125  in  16ths. 
.15625  in  32ds 
.77  in  64ths. 
.81  in  48ths. 


Ans. 


'(«) 

(b) 

(c) 
(■ d ) 


5 

1  6 
5 

32 

49 

64 


(e) 


39 

48 


.923  in  96ths. 


(0 


89 
9  6 


62.  The  sign  for  dollars  is  $.  It  is  read  dollars.  $25  is 
read  25  dollars. 

Since  there  are  100  cents  in  a  dollar,  1  cent  is  1  one- 
hundredth  of  a  dollar;  the  first  two  figures  of  a  decimal  part 
of  a  dollar  represent  cents.  Since  a  mill  is  iV  of  a  cent,  or 
ToVo  of  a  dollar,  the  third  figure  represents  mills. 

Thus,  $25.16  is  read  twenty-five  dollars  and  sixteen  cents; 
$25,168  is  read  twenty-five  dollars  sixteen  cents  and  eight  mills. 


■ 


ARITHMETIC 

(PART  3) 


PERCENTAGE 


DEFINITIONS  AND  PRINCIPLES 

1.  In  certain  operations  it  is  very  convenient  to  regard  a 

quantity  as  being  divided  into  100  equal  parts;  thus,  instead 
of  using  the  ordinary  fractions  i,  f,  f,  the  equivalent  frac¬ 
tions  or  their  equivalent  decimals  .25,  .60,  .28-7 

100  100  100 

are  used.  This  practice  is  a  very  convenient  one  in  all  com¬ 
putations  involving  United  States  money,  because,  since  $1 
equals  100  cents,  it  is  easier  to  comprehend  what  part  of  the 
whole  i^o  is  than  some  other  equivalent  fraction,  as  t4^-;  it 
is  also  much  easier  to  compute  with  fractions  whose  denom¬ 
inators  are  100  than  it  is  to  compute  with  fractions  whose 
denominators  are  composed  of  other  figures. 

2.  Percentage  is  a  term  applied  to  those  arithmetical 
operations  in  which  the  number  or  quantity  to  be  operated  on 

is  supposed  to  be  divided  into  100  equal  parts. 

/ 

3.  The  term  per  cent,  means  by  the  hundred.  Thus, 
8  per  cent,  of  a  number  means  8  hundredths;  i.  e.,  tTo,  or 
.08,  of  that  number;  8  per  cent,  of  250  is  250  X  tTo,  or  250 
X  .08  =  20;  47  per  cent,  of  75  bushels  is  75  X  to  =  75 
X  .47  =  35.25  bushels.  The  statement  that  the  population 
of  a  city  has  increased  22  per  cent,  in  a  given  time,  say  from 


For  notice  of  copyright ,  see  page  immediately  following  the  title  page 

23 


2 


ARITHMETIC 


§3 


1890  to  1900,  is  equivalent  to  saying  that  the  increase  is  22 
in  every  hundred;  that  is,  for  every  100  in  1890,  there  are 
22  more,  or  122,  in  1900. 

4.  The  sign  of  per  cent,  is  %,  and  is  read  per  cent. 
Thus,  6%  is  read  six  per  cent.;  12i%  is  read  twelve  and  one- 
half  per  cent.,  etc. 

5.  When  expressing  the  per  cent,  of  a  number  to  use  in 
calculations,  it  is  customary  to  express  it  decimally  instead 
of  fractionally.  Thus,  instead  of  expressing  6%,  25%, 
and  43%  as  tot,  i^o,  and  it  is  usual  to  express  them 
as  .06,  .25,  and  .43. 

6.  The  following  table  will  show  how  per  cent,  can  be 
expressed  either  as  a  decimal  or  as  a  fraction: 


Per  Cent. 

Decimal 

Fraction 

Per  Cent. 

Decimal 

Fraction 

1%  .  . 

.01 

1 

100 

i%.  . 

.0025 

i  _JL_ 

10  0  OT  4  0  0 

2%  .  . 

.02 

2 

100 

or  bV 

\°fo.  . 

.005 

i  r,r  .  1... 
100  °f  2  00 

s%  .  . 

•05 

5 

100 

or  to 

ii%  .  . 

.015 

li  nr  3 
100  01  200 

10%  .  . 

.IO 

1  0 
10  0 

or  to 

6i%  .'  . 

■  .064 

nr  _1_ 
100  or  1 6 

25%  •  • 

.25 

2  B 

10  0 

or  i 

8i%  .  . 

.083 

8i  1 

Too  or  IT 

50%  .  . 

•50 

B  0 

10  0 

or  2 

122%  .  . 

.125 

I2i  nr  J- 

100  or  8 

75%  •  • 

•75 

7  B 

1  0  0 

or  f 

i6f%.  . 

.i6f 

163  X 

100  or  6 

100%  .  . 

I. OO 

1  0  0 
10  0 

or  1 

33a%  .  . 

•  333 

33*  X 

100  or  3 

125%  .  . 

1.25 

1  2  B 

10  0 

or  1  i 

37*%  •  • 

•  372 

37i  a 

100  or  8 

150%  .  . 

1.50 

1  B  0 
10  0 

or  ii 

62!%  .  . 

.625 

62*  nr  £ 

100  or  8 

500%  .  . 

5.00 

5  0  0 
10  0 

or  5 

87¥%  .  . 

.875 

87*  nr  i 

1 0  0  or  8 

7.  The  names  of  the  terms  used  in  percentage  are: 
the  base ,  the  rate  or  rate  Per  cent.,  the  percentage ,  the  amount , 
and  the  difference. 

8.  The  base  is  the  number  or  quantity  that  is  supposed 
to  be  divided  into  100  equal  parts. 


§3 


ARITHMETIC 


3 


9.  The  rate  per  cent,  is  that  number  of  the  100  equal 

parts  into  which  the  base  is  supposed  to  be  divided  that  is 
taken  or  considered.  The  rate  is  the  number  of  hundredths 
of  the  base  that  is  taken  or  considered.  The  distinction 
between  the  rate  per  cent,  and  the  rate  is  this:  the  rate  per 
cent,  is  always  100  times  the  rate.  Thus,  7%  of  125  and  .07 
of  125  amount  in  the  end  to  the  same  thing;  the  former,  7, 
is  the  rate  per  cent. — the  number  of  hundredths  of  125 
intended;  the  latter,  .07,  is  the  rate,  the  part  of  125  that  is 
to  be  found;  7%  is  used  in  speech,  .07  is  the  form  used  in 
computation.  So,  also,  12 =  .125,  =  .005,  If °Io  =  .0175. 

10.  The  percentage  is  the  result  obtained  by  multiply¬ 
ing  the  base  by  the  rate.  Thus,  7%  of  125  =  125  X  .07 
=  8.75,  the  percentage. 

11.  The  amount  is  the  sum  of  the  base  and  the  per¬ 
centage. 

12.  The  difference  is  the  remainder  obtained  when  the 
percentage  is  subtracted  from  the  base. 

13.  The  terms  amount  and  difference  are  ordinarily  used 
when  there  is  an  increase  or  a  decrease  in  the  base.  For 
example,  suppose  the  population  of  a  village  is  1,500  and  it 
increases  25  per  cent.  This  means  that  for  every  100  of  the 
original  1,500  there  is  an  increase  of  25,  or  a  total  increase  of 
15  X  25  =  375.  This  increase  added  to  the  original  population 
gives  the  amount ,  or  the  population  after  the  increase.  If  the 
population  had  decreased  375,  the  final  population  would 
have  been  1,500  —  375  =  1,125,  and  this  would  be  the  differ¬ 
ence.  The  original  population,  1,500,  is  the  base  on  which 
the  percentage  is  computed;  the  25  is  the  rate  per  cent.,  and 
the  increase  or  decrease,  375,  is  the  percentage.  If  the  base 
increases,  the  final  value  is  the  amount,  and  if  it  decreases, 
its  final  value  is  the  difference. 


4 


ARITHMETIC 


3 


BASE,  RATE,  AND  PERCENTAGE 

14.  Rule. — To  find,  the  percentage ,  multiply  the  base  by 
the  rate. 

Example  1. — A  farmer  raised  650  bushels  of  wheat  and  sold  64%; 
how  many  bushels  did  he  sell? 

Solution. — The  base  is  650  bu.  Out  of  every  100  bu.  raised  64  were 
sold;  that  is,  the  number  of  bushels  sold  was  nf0  or  .64  of  the  number 
raised.  The  rate  is,  then,  .64,  and 

650  X  .64  =  416  bu.,  the  percentage.  Ans. 

Example  2. — If  cotton  is  selling  at  8  cents  per  pound  and  the  price 
goes  up  121%,  what  will  be  the  new  selling  price? 

Solution. — It  is  evident  that  the  new  selling  price  is  equal  to  the 
original  selling  price  plus  the  percentage  of  increase.  The  base  is 
8  ct.;  the  rate  is  Vl\%  ,  or  .125,  and  the  percentage  is  .125  X  8  ct.  =  let. 
Hence,  the  new  selling  price  is  8  ct.  -f  1  ct.  =  9  ct.  Ans. 

15.  In  the  last  example,  the  result  sought,  9  cents,  was 
the  amount.  In  all  cases  where  the  amount  is  involved,  it 
can  be  obtained  directly  by  multiplying  the  base  by  1  plus 
the  rate.  Thus,  in  the  last  example  1  plus  the  rate  is 
1  +  .125  =  1.125,  and  1.125  X  8  cents  =  9  cents,  the  same 
result  as  before. 

Similarly,  the  difference  may  be  found  directly  by  multi- 

% 

plying  the  base  by  1  minus  the  rate.  An  example  will 
illustrate  this. 

Example. — If  wool  is  selling  for  20  cents  per  pound  and  the  price 
drops  10%,  what  will  be  the  new  selling  price? 

Solution. — The  new  selling  price  is  evidently  equal  to  the  original 
selling  price  less  the  percentage  of  decrease.  Here  20  ct.  is  the  base 
the  rate  .10,  and  the  percentage  is  .10  X  20  ct.  =  2  ct.  Hence,  the 
new  selling  price  is  20  ct.  —  2  ct.  =  18  ct.  Ans. 

Or,  the  difference  expressed  as  a  per  cent,  is  1  —  .10  =  .90,  and 
.90  X  20  ct.  =  18  ct.  Ans. 

16.  When  the  percentage  and  rate  are  known,  the  base 
may  be  found  by  dividing  the  percentage  by  the  rate.  Thus, 
suppose  that  12  is  6%,  or  too,  of  some  number;  then  1%,  or 
Too',  of  the  number,  is  12  -t-  6,  or  2.  Consequently,  if  2  =  1% 
or  too-,  100%,  or  too  =  2  X  100  =  200.  But,  since  the  same 


§3 


ARITHMETIC 


5 


result  may  be  arrived  at  by  dividing  12  by  .06,  for  12  -r-  .06 
=  200,  it  follows  that: 

Rule. —  When  the  percentage  and  rate  are  known,  to  find  the 
base ,  divide  the  percentage  by  the  rate ,  expressed  decimally. 

Example  1. — Bought  a  certain  number  of  bushels  of  apples  and 
sold  76%  of  them;  if  I  sold  228  bushels,  how  many  bushels  did  I  buy? 

Solution. — Here  228  is  the  percentage,  and  76%,  or  .76,  is  the  rate; 
hence,  applying  the  rule, 

228  .76  =  300  bu.  Ans. 

Example  2. — A  corporation  is  operating  500  looms,  which  is  25% 
of  the  number  of  looms  in  the  mill;  how  many  looms  are  in  the  mill? 

Solution. — Here  500  is  25%  of  some  number;  in  other  words,  500 
is  the  percentage  and  .25  is  the  rate.  Therefore,  applying  the  rule, 
the  number  of  looms  in  the  mill  equals  the  base,  which  equals  500 
-p  .25  =  2,000.  Ans. 

Example  3. — A  mill  sells  cloth  for  12  cents  per  yard,  which  is  20% 
more  than  the  cost  to  manufacture;  what  is  the  cost  per  yard  to 
manufacture? 

Solution. — The  cost  to  manufacture  is  the  base,  or  100%.  The 
selling  price  is  20%  greater,  or  100%  +  20%  =  120%.  The  rate  is 
therefore  1.20.  Applying  the  rule, 

12  ct.  -r-  1.20  =  10  ct.  Ans. 

Example  4. — Suppose  that  in  the  last  example,  the  selling  price, 
12  cents,  is  20%  less  than  the  cost  to  manufacture;  what  is  the  cost  to 
manufacture? 

Solution. — The  cost  to  manufacture  is  the  base,  12  ct.  is  the  per¬ 
centage,  and  1  —  .20  =  .80  is  the  rate.  Applying  the  rule, 

12  ct.  -r-  .80  =  15  ct.  Ans. 

17.  When  the  base  and  percentage  are  given,  the  rate 
(expressed  decimally)  may  be  found  by  dividing  the  per¬ 
centage  by  the  base.  Thus,  suppose  that  it  is  desired 
to  find  what  per  cent.  12  is  of  200.  1%  of  200  is  200  X  .01 

=  2.  Now,  if  1%  is  2,  12  is  evidently  as  many  per  cent,  as 
the  number  of  times  that  2  is  contained  in  12,  or  12  -4-  2  =  6%. 
But  the  same  result  maybe  obtained  by  dividing  12,  the  per¬ 
centage,  by  200,  the  base,  since  12  4-  200  =  .06  =  6%.  Hence, 

Rule. —  When  the  percentage  and  base  are  given,  to  find  the 
rate ,  divide  the  percentage  by  the  base ,  and  the  result  will  be  the 
rate ,  expressed  decimally. 


6  ARITHMETIC  §3 

Example  1. — Bought  300  bushels  of  apples  and  sold  228  bushels. 
What  per  cent,  of  the  total  number  of  bushels  was  sold? 

Solution. — Here  300  is  the  base  and  228  is  the  percentage;  hence, 
applying  rule, 

rate  =  228  -^  300  =  .76  =  76%.  Ans. 

Example  2. — What  per  cent,  of  875  is  25? 

Solution. — Here  875  is  the  base,  and  25  is  the  percentage;  hence, 
applying  rule, 

25  -T-  875  =  .02f  =  2f%.  Ans. 

Example  3. — A  mill  contains  300  looms,  which  are  capable  of  pro¬ 
ducing  200  cuts  per  day  of  a  certain  kind  of  cloth,  if  running  all  the 
time.  The  actual  production  is  180  cuts  per  day.  What  per  cent,  of 
the  possible  total  production  per  day  are  the  looms  turning  off? 

Solution. — Here  200  is  the  base,  because  the  question  is,  What  per 
cent,  of  the  total  production  is  180?  Hence,  180  is  the  percentage  and 
200  is  the  base.  Therefore,  applying  the  rule, 

180  -f-  200  =  .90  =  90%.  Ans. 

Example  4. — Cloth  costing  75  cents  per  yard  to  manufacture  is  sold 
at  60  cents  per  yard;  what  is  the  loss  per  cent.? 

Solution. — Here  75  ct.  is  the  base,  because  it  is  desired  to  know 
the  loss  per  cent,  on  the  cost.  The  loss  was  75  ct.  —  60  ct.  =  15  ct. 
Hence, 

15  ct.  -5-  75  ct.  =  .20  =  20%.  Ans. 

Example  5. — Suppose  that,  in  the  last  example,  the  cloth  had  been 
sold  for  80  cents  per  yard;  what  would  have  been  the  gain  per  cent.? 

Solution. — As  before,  75  ct.  is  the  base;  the  gain  is  80  ct.  —  75  ct. 
=  5  ct.;  and  the  gain  per  cent,  is  5ct.  -5-  75  ct.  =  ,06f  =  6|%.  Ans. 


EXAMPLES  FOR  PRACTICE 

1.  Cloth  costing  30  cents  per  yard  to  manufacture  was  sold  at  a 

loss  of  16|%;  what  was  the  selling  price  per  yard?  Ans.  25  ct. 

2.  Cloth  costing  60  cents  per  yard  to  manufacture  is  sold  at  70 

cents  per  yard;  what  is  the  gain  per  cent.?  Ans.  16f% 

3.  Cloth  is  sold  at  70  cents  per  yard,  which  gives  a  profit  of  20%; 

what  is  the  cost  per  yard  to  manufacture?  Ans.  58£  ct. 

4.  An  agent  receives  $50  for  selling  a  machine.  The  money  he 

receives  is  20%  of  the  cost  of  the  machine;  what  is  the  cost  of  the 
machine?  Ans.  $250 


§3 


ARITHMETIC 


7 


5.  A  certain  amount  of  yarn  weighs  1,800  pounds  before  being 

sized,  and  2,000  pounds  after  being  sized;  what  per  cent,  of  this  2,000 
pounds  is  yarn,  and  what  per  cent,  is  size?  Ang  j 90%  yarn 

’\10%  size 

6.  A  mixing  of  wool  has  to  weigh  730  pounds,  of  which  9%  is 

red,  17%  black,  69%  natural,  and  the  balance  olive;  find  the  weight 
of  each  color  required.  165.7  lb.  red 

A  124.1  lb.  black 
'503.71b.  natural 
36.5  lb.  olive 


7.  A  cotton  broker  sold  100  bales  of  cotton  for  $4,000,  and  charged 
$200  commission;  what  rate  per  cent,  did  he  charge  for  selling? 

Ans.  5% 

8.  The  production  of  a  machine  has  to  be  increased  8%.  It  now 
turns  off  14  ounces  per  spindle  in  a  given  time;  how  many  ounces 
will  a  spindle  produce  in  the  same  time  after  the  change? 

Ans.  15.12  oz. 


9.  A  comber  turns  off  50  pounds  of  cotton  in  a  day,  which  is  84 
per  cent,  of  the  cotton  fed  in:  (a)  What  is  the  weight  of  the  cotton  fed 
in?  (A)  How  many  pounds  of  waste  are  there  in  one  day? 


Ans. 


{ 


(a)  59.51b. 

(b)  9.51b. 


INTEREST 

18.  Interest  as  a  general  term  may  be  defined  as  a 
form  of  percentage  dealing  particularly  with  the  use  of  money. 
The  rules  and  explanations  previously  given  will  be  found 
to  apply  equally  well  when  dealing  with  interest. 

One  point,  however,  that  should  be  carefully  noted  is  that 
whenever  the  use  of  money  is  being  considered  the  element 
of  time  enters  into  the  calculation. 

19.  The  terms  commonly  used  in  connection  with  this 
subject  are  as  follows: 

Interest,  which  is  the  sum  of  money  paid  for  the  use  of 
money  for  a  certain  time. 

Principal,  which  is  the  money  for  the  use  of  which  the 
interest  is  paid. 

Rate  of  interest,  which  is  a  certain  per  cent,  and  which  is 
also  known  as  the  rate  per  cent. 

Amount,  which  is  the  principal  and  the  interest  added 
together. 


8 


ARITHMETIC 


3 


20.  To  illustrate  these  terms  suppose  that  a  person  bor¬ 
rows  $100  for  a  year,  at  the  end  of  which  time  he  pays 
the  $100  and  also  the  interest  at  the  rate  of  6%  per  year. 
6%  of  $100  is  $6.  Therefore,  to  cancel  his  debt  he  will 
pay  $100  +  $6,  or  $106. 

In  the  above  case  $100  is  the  principal ,  or  the  money 
loaned  by  one  person  and  borrowed  by  another,  $6  is  the 
interest ,  or  the  amount  of  money  the  borrower  pays  for  the 
use  of  the  principal,  6%  per  year  is  the  rate  per  cent.,  and 
$106  is  the  amount. 

21.  The  unit  of  time  when  figuring  interest  is  generally 
one  year.  Very  frequently  the  time  is  not  stated  and  in  all 
such  cases  should  be  considered  as  one  year. 

Example.— Find  the  interest  on  $500  for  4  years  at  2% . 

Solution. — In  such  a  case  the  time  is  understood  to  be  one  year; 
that  is,  2%  per  year.  Thus, 

$  5  0  0 
.0  2 

$  1  0.0  0  int.  for  1  yr. 

$10  X  4  =  $40  int.  for  4  yr.  Ans. 

22.  Interest,  however,  may  be  computed  for  shorter 
periods,  as  semiannually  (6  mo.)  or  quarterly  (3  mo.), 
although  the  general  rule  is  1  year. 

Example. — A  man  borrows  $400  for  1  year  agreeing  to  pay 
2%  quarterly;  how  much  interest  will  he  have  paid  at  the  end  of 
the  year? 

Solution. —  $  4  0  0 

.0  2 

$  8.0  0  int.  for  3  mo. 

$8X4  =  $32  int.  for  1  yr.  Ans. 

23.  Another  form  of  interest  is  that  known  as  compound 
interest.  In  this  case  the  interest  is  added  to  the  principal 
at  certain  intervals  and  this  amount  is  taken  as  a  new 
principal. 

Interest  may  be  compounded,  or  added  to  the  principal, 
annually,  semiannually,  or  quarterly,  but  if  the  time  is  not 
stated  it  is  understood  to  be  1  year. 


§3 


ARITHMETIC 


9 


Example. — Find  the  amount  to  be  paid  at  the  end  of  1  year 
6  months  on  $500  at  4%  compounded  semiannually. 


Solution. — In  this  example,  the  rate  4%  is  for  1  yr.,  but  the 
interest  must  be  added  at  the  end  of  every  6  mo.;  consequently,  the 
rate  must  be  considered  as  2%  every  6  mo.,  which  equals  4%  for 
1  yr.  The  method  of  performing  the  example  is  as  follows: 


$500 
.0  2 

$  1  0.0  0 
5  0  0 
$  5  1  0 
.0  2 

$  1  0.2  0 
5  1  0 

$  5  2  0.2  0 
.0  2 

$  1  0.4  0  4  0 
5  2  0.2  0 

$  5  3  0.6  0 


int.  at  end  of  6  mo. 
first  principal 
new  principal 

int.  at  end  of  1  yr. 
second  principal 

new  principal 

int.  at  end  of  1  yr.  6  mo. 
third  principal 

amount  at  end  of  1  yr.  6  mo.  Ans. 


EXAMPLES  FOR  PRACTICE 

1.  What  is  the  interest  on  $800  for  5  years  at  6%?  Ans.  $240 

2.  What  is  the  amount  to  be  paid  at  the  end  of  3  years  for  $300 

at  4%  ?  Ans.  $336 

3.  A  person  pays  $40  per  year  for  the  use  of  a  certain  sum  of 
money,  the  interest  being  4%;  what  is  the  principal?  Ans.  $1,000 

4.  If  $36  is  paid  for  2  years’  interest  on  a  sum  of  money,  the  rate 

per  cent,  being  3,  what  is  the  principal?  Ans.  $600 

5.  What  is  6  months’  interest  on  $5,000  at  4%.  Ans.  $100 


10 


ARITHMETIC 


§3 


RATIO 

24.  Suppose  that  it  is  desired  to  compare  two  numbers, 
say  20  and  4.  If  we  wish  to  know  how  many  times  larger 
20  is  than  4,  we  divide  20  by  4  and  obtain  5  for  the  quotient; 
thus,  20  -T-  4  =  5.  Hence,  we  say  that  20  is  5  times  as  large 
as  4;  i.  e.,  20  contains  5  times  as  many. units  as  4.  Again, 
suppose  that  we  desire  to  know  what  part  of  20  is  4.  We 
then  divide  4  by  20  and  obtain  i;  thus,  4  -f-  20  =  i,  or  .2. 
Hence,  4  is  i,  or  .2,  of  20.  This  operation  of  comparing  two 
numbers  is  termed  finding  the  ratio  of  the  two  numbers. 
Ratio,  then,  is  a  comparison.  It  is  evident  that  the  two 
numbers  to  be  compared  must  be  expressed  in  the  same  unit; 
in  other  words,  the  two  numbers  must  both  be  abstract  num- 
bers  or  concrete  numbers  of  the  same  kind.  For  example, 
it  would  be  absurd  to  compare  20  horses  with  4  birds,  or 
20  horses  with  4.  Hence,  ratio  may  be  defined  as  a  com¬ 
parison  between  two  numbers  of  the  same  kind. 

25.  .  A  ratio  may  be  expressed  in  three  ways;  thus,  if  it  is 
desired  to  compare  20  and  4  and  express  this  comparison  as 

20 

a  ratio,  it  may  be  done  as  follows:  20  -f-  4;  20  :  4,  or  - — .  All 

4 

three  express  the  ratio  of  20  to  4.  The  ratio  of  4  to  20  would 

4 

be  expressed  thus:  4  -4-  20;  4  :  20,  or  — .  The  first  method  of 

expressing  a  ratio,  although  correct,  is  seldom  or  never 
used;  the  second  form  is  the  one  most  often  met  with,  while 
the  third  form,  called  the  fractional  form,  possesses  great 
advantages  to  students  of  algebra  and  of  higher  mathe¬ 
matical  subjects.  The  second  form  is  better  adapted  to 
arithmetical  subjects  and  is  the  one  ordinarily  adopted  in 
this  Course. 

26.  The  terms  of  a  ratio  are  the  two  numbers  to  be 
compared;  thus,  in  the  above  ratio,  20  and  4  are  the  terms. 


§3 


ARITHMETIC 


11 


When  both  terms  are  considered  together,  they  are  called  a 
couplet;  when  considered  separately,  the  first  term  is  called 
the  antecedent  and  the  second  term,  the  consequent. 
Thus,  in  the  ratio  20  :  4,  20  and  4  form  a  couplet,  and  20  is 
the  antecedent  and  4  the  consequent. 

27.  A  ratio  may  be  direct  or  inverse.  The  direct  ratio 
of  20  to  4  is  20  :  4,  while  the  inverse  ratio  of  20  to  4  is  4  :  20. 
The  direct  ratio  of  4  to  20  is  4  :  20,  and  the  inverse  ratio  is 
20  :  4.  An  inverse  ratio  is  sometimes  called  a  reciprocal 
ratio.  The  reciprocal  of  a  number  is  1  divided  by  the 
number.  Thus,  the  reciprocal  of  17  is  iV;  of  f  is  1  -r-  f  =  I; 
i.  e.,  the  reciprocal  of  a  fraction  is  the  fraction  inverted. 
Hence,  the  inverse  ratio  of  20  to  4  may  be  expressed  as  4  :  20, 
or  as  to  :  i.  Both  have  equal  values;  for,  4  -f-  20  =  i,  and 

_J_.i__l_w4._A  % 

20  “  4  —  2  0  X  1  —  5  • 

28.  The  term  vary  implies  a  ratio.  When  we  say  that 
two  numbers  vary  as  some  other  two  numbers,  we  mean  that 
the  ratio  between  the  first  two  numbers  is  the  same  as  the 
ratio  between  the  other  two  numbers. 

29.  The  value  of  a  ratio  is  the  result  obtained  by  per¬ 
forming  the  division  indicated.  Thus,  the  value  of  the  ratio 
20  :  4  is  5;  it  is  the  quotient  obtained  by  dividing  the  ante¬ 
cedent  by  the  consequent.  The  value  of  a  ratio  is  always 
an  abstract  number,  regardless  of  whether  the  terms  are 
abstract  or  concrete  numbers. 

30.  When  a  ratio  is  expressed  in  words,  as  the  ratio  of 
20  to  4,  the  first  number  named  is  always  regarded  as  the 
antecedent  and  the  second  as  the  consequent,  without  regard 
to  whether  the  ratio  itself  is  direct  or  inverse.  When  not 
otherwise  specified ,  all  ratios  are  understood  to  be  direct.  To 
express  an  inverse  ratio,  the  simplest  way  of  doing  it  is  to 
express  it  as  if  it  were  a  direct  ratio,  with  the  first  number 
named  as  the  antecedent,  and  then  transpose  the  antecedent 
to  the  place  occupied  by  the  consequent,  and  the  consequent 
to  the  place  occupied  by  the  antecedent;  or,  if  expressed  in 
the  fractional  form,  invert  the  fraction.  Thus,  to  express  the 


12 


ARITHMETIC 


3 


inverse  ratio  of  20  to  4,  first  write  it  20  :  4,  and  then,  trans¬ 
posing  the  terms,  as  4  :  20;  or  as  •Sia,  and  then  inverting, 
as  2*d.  Or,  the  reciprocals  of  the  numbers  may  be  taken,  as 
explained  above.  To  invert  a  ratio  is  to  transpose  its  terms. 


31.  Instead  of  expressing  the  value  of  a  ratio  by  a  single 
number,  it  is  more  convenient  to  express  it  by  means  of 
another  ratio  in  which  the  consequent  is  1.  Thus,  suppose 
that  it  is  desired  to  find  the  ratio  of  the  weights  of  two  pieces 
of  iron,  one  weighing  45  pounds  and  the  other  weighing 
30  pounds.  The  ratio  of  the  heavier  to  the  lighter  is  then 
45  :  30,  an  inconvenient  expression.  Using  the  fractional 


form,  we  have 


45 


30 

quent,  we  obtain 


U 


Dividing  both  terms  by  30,*  the  conse- 
,  or  I2  :  1.  This  is  the  same  result  as 


obtained  above,  for  li  -f-  1  =  li,  and  45  -f-  30  =  I2. 


PROPORTION 

32.  Proportion  is  an  equality  of  ratios,  the  equality 
being  indicated  by  the  double  colon  (  :;  )  or  by  the  sign  of 
equality  (  =  ).  Thus,  to  write  in  the  form  of  a  proportion  the 
two  equal  ratios,  8  :  4  and  6  :  3,  which  both  have  the  same 
value,  2,  we  may  employ  any  one  of  the  three  following  forms: 

8  :  4  ::  6  :  3  (1) 

8  :  4  =  6  :  3  (2) 


33.  The  first  form  is  the  one  most  extensively  used,  by 
reason  of  its  having  been  exclusively  employed  in  all  the 
older  works  on  mathematics.  The  second  and  third  forms 
are  being  adopted  by  all  modern  writers  on  mathematical 
subjects,  and  in  time  will  probably  entirely  supersede  the 
first  form.  In  this  Course,  the  second  form  will  be  adopted, 

*This  evidently  does  not  alter  the  value  of  the  ratio,  since  by  the 
laws  or  fractions,  both  numerator  and  denominator  may  be  divided  by 
the  same  number  without  changing  the  value  of  the  fraction. 


§3 


ARITHMETIC 


13 


unless  some  statement  can  be  made  clearer  by  using  the 
third  form. 

34.  A  proportion  may  read  in  two  ways.  The  old 
way  to  read  the  preceding  proportion  was:  8  is  to  A  as  6  is  to  3; 
the  new  way  is:  the  ratio  of  8  to  4  equals  the  ratio  of  6  to  3. 
The  student  may  read  it  either  way,  but  we  recommend  the 
latter. 

35.  Each  ratio  of  a  proportion  is  termed  a  couplet.  In 
the  above  proportion,  8  :  4  is  a  couplet,  and  so  is  6  :  3. 

36.  The  numbers  forming  the  proportion  are  called 
terms;  and  they  are  numbered  consecutively  from  left  to 
right,  thus: 

first  second  third  fourth 

8  :  4.=  6  :  3 

Hence,  in  any  proportion,  the  ratio  of  the  first  term  to  the 
second  term  equals  the  ratio  of  the  third  term  to  the  fourth 
term. 

37.  The  first  and  fourth  terms  of  a  proportion  are  called 
the  extremes,  and  the  second  and  third  terms  the  means. 
Thus,  in  the  foregoing  proportion,  8  and  3  are  the  extremes 
and  4  and  6  are  the  means. 

38.  A  direct  proportion  is  one  in  which  both  couplets 
are  direct  ratios. 

39.  An  inverse  proportion  is  one  that  requires  one 
of  the  couplets  to  be  expressed  as  an  inverse  ratio.  Thus, 
8  is  to  4  inversely  as  3  is  to  6  must  be  written  8  :  4  =  6  :  3; 
i.  e.,  the  second  ratio  (couplet)  must  be  inverted. 

40.  Proportion  forms  one  of  the  most  useful  sections  of 
arithmetic.  In  our  grandfathers’  arithmetics,  it  was  called 
“The  rule  of  three.” 

41.  Rule. — In  any  proportion ,  the  product  of  the  extremes 
equals  the  product  of  the  means. 

Thus,  in  the  proportion 

17  :  51  =  14  :  42 

17  X  42  =  51  X  14,  since  both  products  equal  714. 


14  ARITHMETIC  §3 


42.  Rule. — The  product  of  the  extremes  divided  by  either 
mean  gives  the  other  mean. 

Example. — What  is  the  third  term  of  the  proportion  17  :  51  =  :  42? 

Solution. — Applying  the  rule, 

17  X  42  =  714,  and  714  -r  51  =  14.  Ans. 

43.  Rule. — The  product  of  the  means  divided  by  either 
extreme  gives  the  other  extreme. 

Example. — What  is  the  first  term  of  the  proportion  :  51  =  14  :  42? 

Solution. — Applying  the  rule, 

51  X  14  =  714,  and  714  -f-  42  =  17.  Ans. 

44.  When  stating  a  proportion  in  which  one  of  the  terms 
is  unknown,  represent  the  missing  term  by  a  letter,  as  x. 
Thus,  the  last  example  would  be  written 

x  :  51  =  14  :  42 

and  for  the  value  of  x,  we  have  x  =  ^  ^  =  17. 


45.  The  principle  of  all  calculations  in  proportion  is 
this:  Three  of  the  ter?ns  are  always  given  a?id  the  remaining 
one  is  to  be  fou?id. 

Example  1. — If  4  men  can  earn  $25  in  1  week,  how  much  can 
12  men  earn  in  the  same  time? 

Solution. — The  required  term  must  bear  the  same  relation  to  the 
given  term  of  the  same  kind  as  one  of  the  remaining  terms  bears  to 
the  other  remaining  term.  We  can  then  form  a  proportion  by  which 
the  required  term  may  be  found. 

The  first  question  the  student  must  ask  himself  in  every  calculation 
in  proportion  is:  “What  is  it  I  want  to  find?”  In  this  case  it  is 
dollars.  We  have  two  sets  of  men,  one  set  earning  $25,  and  we  want 
to  know  how  many  dollars  the  other  set  earns.  It  is  evident  that  the 
amount  12  men  earn  bears  the  same  relation  to  the  amount  that  4  men 
earn  as  12  men  bears  to  4  men.  Hence,  we  have  the  proportion,  the 
amount  12  men  earn  is  to  $25  as  12  men  is  to  4  men;  or,  x  :  $25  =  12 
men  :  4  men.  Since  either  extreme  equals  the  product  of  the  means 
divided  by  the  other  extreme,  we  have, 


Since  it  matters  not  which  place  x,  or  the  required  term,  occupies, 
the  problem  could  be  stated  in  any  of  the  following  forms,  the  value  of 
x  being  the  same  in  each: 


3 


ARITHMETIC 


15 


(a)  $25  :  x  =  4  men  :  12  men;  or  the  amount  12  men  earn 
$25  X  12 

=  - 1 or  $75,  since  either  mean  equals  the  product  of  the 

extremes  divided  by  the  other  mean. 

(b)  4  men  :  12  men  =  $25  :  x\  or  the  amount  that  12  men  earn 
$25  X  12 

=  - - - ,  or  $75,  since  either  extreme  equals  the  product  of  the 

means  divided  by  the  other  extreme. 

t 

( c )  12  men  :  4  men  =  x  :  $25;  or  the  amount  that  12  men  earn 
$25  X  12 

=  - - - ,  or  $75,  since  either  mean  equals  the  product  of  the 

extremes  divided  by  the  other  mean. 


Example  2. — If  200  looms  produce  300  cuts  in  a  given  time,  how 
many  cuts  will  400  looms  produce  in  the  same  time,  when  working 
under  similar  conditions? 


Solution. — Comparing  the  number  of  looms  in  the  two  cases,  the 
ratio  200  :  400  is  obtained;  comparing  the  number  of  cuts,  the  ratio  is 
300  :  x.  Hence,  the  proportion  is 

200  :  400  =  300  :  x 
400  X  300 


from  which 


x  = 


200 


=  600  cuts.  Ans. 


Another  way  of  stating-  the  proportion  in  examples  like 
the  one  last  given  is  to  write  the  first  number,  in  this  case 
200  looms;  then  write  the  next  number  of  the  same  kind,  in 
this  case  400  looms,  with  the  ratio  sign  between;  write  the 
sign  of  equality  and  then  the  number  mentioned  in  connec¬ 
tion  with  the  first  and  which  results  from  it,  in  this  case 
300  cuts;  lastly,  write  the  number  of  the  same  kind  as  the 
third  number.  An  example  will  illustrate  this. 

Example  3. — If  60  bales  of  cotton  are  used  in  5  days,  how  many 
bales  will  be  used  in  18  days  at  the  same  rate? 

Solution. — The  first  number  is  60  bales;  the  other  number 
expressed  in  bales  is  to  be  found  and  is  represented  by  x.  The  num¬ 
ber  mentioned  in  connection  with  the  first  and  resulting  from  it  is 
5  da.;  that  is,  it  took  5  da.  to  use  60  bales;  hence,  5  da.  is  the  third 
term  and  18  da.  is  the  last,  or  fourth,  term.  The  proportion  is,  therefore, 
60  bales  :  x  bales  =  5  da.  :  18  da. 
or,  60  :  x  =  5  :  18 

from  which  x  =  ^  ^5 — -  =  216  bales.  Ans. 


16 


ARITHMETIC 


3 


EXAMPLES  FOR  PRACTICE 
Find  the  value  of  x  in  each  of  the  following: 


(a) 

$16 

:  $64 

x  :  $4. 

(a) 

x  = 

$1 

(*) 

x  : 

85  = 

10 

:  17. 

(b) 

X  — 

50 

24 

x  — 

15 

:  40. 

(c) 

X  = 

64 

(d) 

18 

94  = 

2 

x.  Ans.< 

(d) 

X  = 

10| 

(e) 

$75 

:  $100 

— 

x  :  100. 

(e) 

X  = 

75 

(f) 

15 

x  = 

21 

10. 

(0 

X  = 

7y 

Or) 

x  : 

75  yd. 

= 

$15  :  $5. 

(g) 

X  = 

225  yd. 

1.  If  75  pounds  of  lead  costs  $2.10,  what  will  125  pounds  cost  at  the 

same  rate?  Ans.  $3.50 

2.  If  A  does  a  piece  of  work  in  4  days  and  B  does  it  in  7  days,  how 

long  will  it  take  A  to  do  what  B  does  in  63  days?  Ans.  36  da. 

3.  How  long  will  it  take  60  cards  to  card  a  certain  amount  of 

cotton  if  40  cards  can  do  the  work  in  3  days?  Ans.  2  da. 

4.  A  mill  is  running  10  hours  a  day  for  6  days  in  the  week  and  is 

turning  off  2,400  cuts  a  week,  but  they  wish  to  run  on  reduced  time  so 
that  their  production  will  be  but  1,600  cuts  per  week;  how  many  hours 
per  day  will  the  mill  run?  Ans.  6f  hr. 

5.  If  4  combers  turn  off  a  certain  weight  of  cotton  in  6  days,  how 

many  combers  will  have  to  be  procured  to  do  the  same  work  in 
2  days?  Ans.  12  combers 


INVERSE  PROPORTION 

46.  In  Art.  39,  an  inverse  proportion  was  defined  as  one 
that  required  one  of  the  couplets  to  be  expressed  as  an 
inverse  ratio.  Sometimes  the  word  inverse  occurs  in  the 
statement  of  the  example;  in  such  cases  the  proportion  can 
be  written  directly,  merely  inverting  one  of  the  couplets. 
But  it  frequently  happens  that  only  by  carefully  studying 
the  conditions  of  the  example  can  it  be  ascertained  whether 
the  proportion  is  direct  or  inverse.  When  in  doubt,  the 
student  can  always  satisfy  himself  as  to  whether  the  propor¬ 
tion  is  direct  or  inverse  by  first  ascertaining  what  is  required, 
and  stating  the  proportion  as  a  direct  proportion.  Then,  in 
order  that  the  proportion  may  be  true,  if  the  first  term  is 
smaller  than  the  second  term ,  the  third  term  must  be  smaller 
than  the  fourth;  or  if  the  first  term  is  larger  tha?i  the  second 


3 


ARITHMETIC 


17 


term ,  the  third  term  must  be  larger  than  the  fourth  term. 
Keeping  this  'in  mind,  the  student  can  always  tell  whether 
the  required  term  will  be  larger  or  smaller  than  the  other  term 
of  the  couplet  to  which  the  required  term  belongs.  Having 
determined  this,  the  student  then  refers  to  the  example  and 
ascertains  from  its  conditions  whether  the  required  term  is 
to  be  larger  or  smaller  than  the  other  term  of  the  same 
kind.  If  the  two  determinations  agree,  the  proportion  is 
direct;  otherwise,  it  is  inverse,  and  one  of  the  couplets  must 
be  inverted. 

Example. — A’s  rate  of  doing  work  is  to  B’s  as  5  :  7;  if  A  does  a 
piece  of  work  in  42  days,  in  what  time  will  B  do  it? 

Solution. — The  required  term  is  the  number  of  days  it  will  take  B 
to  do  the  work.  Hence,  stating  as  a  direct  proportion, 

5:7  —  42  :  x 

Now,  since  7  is  greater  than  5,  at  will  be  greater  than  42.  But,  refer¬ 
ring  to  the  statement  of  the  example,  it  is  easy  to  see  that  B  works 
faster  than  A;  hence,  it  will  take  B  a  less  number  of  days  to  do  the 
work  than  A.  Therefore,  the  proportion  is  an  inverse  one,  and  should 
be  stated 

5:7  —  x  42 
5  X  42 

from  which  x  =  — = —  -  30  da.  Ans. 

Had  the  example  been  stated  thus:  The  time  that  A  requires  to  do 
a  piece  of  work  is  to  the  time  that  B  requires,  as  5  :  7;  A  can  do  it  in 
42  da.,  in  what  time  can  B  do  it?  it  is  evident  that  it  would  take  B 
a  longer  time  to  do  the  work  than  it  would  A;  hence,  x  would  be 
greater  than  42,  and  the  proportion  would  be  direct,  the  value  of  x  being 

— =■ —  =  58.8  da. 

o 


EXAMPLES  FOR  PRACTICE 

1.  If  a  pump  that  discharges  4  gallons  of  water  per  minute  can  fill 

a  tank  in  20  hours,  how  long  will  it  take  a  pump  discharging  12  gal¬ 
lons  per  minute  to  fill  it?  Ans.  6f  hr. 

2.  The  circular  seam  of  a  boiler  requires  50  rivets  when  the  pitch 
is  2i  inches;  how  many  would  be  required  if  the  pitch  were  3|  inches? 

Ans.  40 

3.  The,  spring  hangers  on  a  certain  locomotive  are  2i  inches  wide 

and  -f  inch  thick;  those  on  another  engine  are  of  the  same  sectional 
area,  but  are  3  inches  wide;  how  thick  are  they?  Ans.  f  in. 


18 


ARITHMETIC 


§3 


4.  A  locomotive  with  driving  wheels  16  feet  in  circumference  runs 
a  certain  distance  in  5,000  revolutions;  how  many  revolutions  would  it 
make  in  going  the  same  distance  if  the  wheels  were  22  feet  in  circum¬ 
ference  (no  allowance  for  slip  being  made  in  either  case)  ? 

Ans.  3,636a  rev. 


UNIT  METHOD 


47.  In  the  older  books  on  arithmetic,  a  large  number  of 
problems  were  solved  by  compound  proportion;  but  these 
problems  can  be  solved  much  more  easily  by  the  unit 
method,  which  will  now  be  explained  by  means  of  examples. 


Example  1. — If  a  pump  discharging  4  gallons  of  water  per  minute 
can  fill  a  tank  in  20  hours,  how  long  will  it  take  a  pump  discharging 
12  gallons  per  minute  to  fill  the  tank? 


Solution.— -A  pump  discharging  4  gal.  per  min.  fills  the  tank 
in  20  hr.  Therefore,  a  pump  discharging  1  gal.  per  min.  fills  it 
in  4  X  20  hr.  Hence,  a  pump  discharging  12  gal.  per  min.  fills  it  in 
4  X  20  hr.  _  20  hr. 

12  "  ~  3 


-  =  6|  hr. 


Ans. 


Example  2. — If  4  men  earn  $65.80  in  7  days,  how  much  can  14  men, 
paid  at  the  same  rate,  earn  in  12  days? 


Solution. — 
Therefore, 

Therefore, 

Therefore, 

Therefore,  14 


4  men  in  7  da.  earn  $65.80 

,  •  -  .  $65.80 

1  man  in  7  da.  earns  — - — 

4 

,  .'  .  .  $65.80 

1  man  in  1  da.  earns  — — 

»  X  » 


1  man  in  12  da.  earns 
men  in  12  da.  earn 


$65.80  X  12 
4X7 
$65.80  X  12  X  14 
4X7 


Canceling, 

14  men  in  12  da.  earn  $65.80  X  3  X  2  =  $65.80  X  6  =  $394.80. 


Ans. 


48.  The  student  will  notice  that  in  the  solution  of  these 
examples,  in  the  successive  steps,  the  operations  of  mul¬ 
tiplication  and  division  were  merely  indicated,  and  no 
multiplication  or  division  was  performed  until  the  very  last, 
and  then  the  answer  was  obtained  easily  by  cancelation. 
In  arithmetical  calculations,  the  student  should  make  it  an 
invariable  habit  to  indicate  the  multiplications  and  divisions 
that  occur  in  the  successive  steps  of  a  solution,  and  not  to 


§3 


ARITHMETIC 


19 


perform  these  operations  until  the  very  last.  Then,  he  will 
probably  be  able  to  use  the  principle  of  cancelation. 

Example. — If  a  block  of  granite  8  feet  long,  5  feet  wide,  and  3  feet 
thick  weighs  7,200  pounds,  what  is  the  weight  of  a  block  of  granite 
12  feet  long,  8  feet  wide,  and  5  feet  thick? 


Solution. — If  a  block  8  ft.  long,  5  ft.  wide,  and  3  ft.  thick  weighs 


7,200  lb.,  a  block  1  ft.  long,  5  ft.  wide,  and  3  ft.  thick  weighs 


7,200 


lb.; 


a  block  1  ft.  long,  1  ft.'  wide,  and  3  ft.  thick  weighs 


7,200 

8X5 


lb.;  and  a  block 


7  200 

1  ft.  long,  1  ft.  wide,  and  1  ft.  thick  weighs  ’  lb.  Therefore, 

8X5  X  o 

by  the  same  reasoning,  a  block  12  ft.  long,  8  ft.  wide,  and  5  ft.  thick 
weighs 


7,200  X  12  X  8  X  5 
8X5X3 


4 

7,200  X  It  X  $  X  5 

8x5x3 


28,800  lb.  Ans. 


EXAMPLES  FOR  PRACTICE 

1.  If  a  pump  discharges  90,000  gallons  of  water  in  20  hours,  in 

what  time  will  it  discharge  144,000  gallons?  Ans.  32  hr. 

2.  If  15  looms  produce  450  yards  of  cloth  in  a  certain  length  of 
time,  how  many  yards  will  75  looms  produce  in  the  same  time? 

Ans.  2,250  yd. 


SIGNS  OF  AGGREGATION 

49.  The  vinculum - ,  parentheses  ( ) ,  brackets  [  ]  , 

and  braces  {  }  are  called  symbols  of  aggregation,  and  are 
used  to  include  numbers  that  are  to  be  considered  together. 
Thus,  13  X  8  —  3,  or  13  X  (8  —  3),  shows  that  3  is  to  be 
taken  from  8  before  multiplying  by  13. 

13  X  8^3  =  13  X  5  =  65 
13  X  (8  -  3)  =  13  X  5  =  65 

When  the  vinculum  or  parentheses  are  not  used,  we  have 
13  X  8  -  3  =  104  -  3  =  101 

50.  In  any  series  of  numbers  connected  by  the  signs  +  , 
—  „  X,  and  -s-,  the  operations  indicated  by  the  signs  must  be 
performed  in  order  from  left  to  right,  except  that  no  addition 


20 


ARITHMETIC 


§3 


or  subtraction  may  be  performed  if  a  sign  of  multiplication 
or  division  follows  the  number  on  the  right  of  a  sign  of  addi¬ 
tion  or  subtraction,  until  the  indicated  multiplication  or 
division  has  been  performed.  In  all  cases  the  sign  of  multi¬ 
plication  takes  the  precedence,  the  reason  being  that  when 
two  or  more  numbers  or  expressions  are  connected  by  the 
sign  of  multiplication,  the  numbers  thus  connected  are 
regarded  as  factors  of  the  product  indicated,  and  not  as 
separate  numbers. 

Example  1. — What  is  the  value  of  4  X  24  —  8  +  17? 

Solution. — Performing  the  operations  in  order  from  left  to  right, 
4  X  24  =  96;  96  —  8  =  88;  88  +  17  =  105.  Ans. 

Example  2. — What  is  the  value  of  the  following  expression:  1,296 
-e-  12  +  160  -  22  X  3i  =  ? 

Solution. —  1,296  4-  12  =  108;  108  +  160  =  268;  here  22  cannot  be 
subtracted  from  268  because  the  sign  of  multiplication  follows  22; 
hence,  multiplying  22  by  31  gives  77,  and  268  —  77  =  191.  Ans. 

Had  the  above  expression  been  written  1,296  4-  12  +  160 
—  22  X  3i  r  7  +  25,  it  would  have  been  necessary  to 
divide  22  X  3|  by  7  before  subtracting,  and  the  final  result 
would  have  been  22  X  3a  =  77;  77  ■—  7  =  11;  268  —  11 
=  257;  257  +  25  =  282.  In  other  words,  it  is  necessary  to 
perform  all  the  indicated  multiplication  or  division  included 
between  the  signs  +  and  — ,  or  —  and  +  ,  before  adding  or 
subtracting.  Also,,  had  the  expression  been  written  1,296 
-4-  12  +  160  —  24|  -4-  7  x  32  +  25,  it  would  have  been  neces¬ 
sary  to  multiply  3i  by  7  before  dividing  241,  since  the  sign 
of  multiplication  takes  the  precedence,  and  the.  final  result 
would  have  been  31  X  7  =  24l;  24l  4-  24l  =  1;  268  —  1 
=  267;  267  +  25  =  292. 

It  likewise  follows  that  if  a  succession  of  multiplication 
and  division  signs  occur,  the  indicated  operations  must  not 
be  performed  in  order,  from  left  to  right — the  multiplication 
must  be  performed  first.  Thus,  24  x34-4x24-9x5  =  1. 
In  order  to  obtain  the  same  result  that  would  be  obtained  by 
performing  the  indicated  operations  in  order,  from  left  to 
right,  symbols  of  aggregation  must  be  used.  Thus,  by  using 


§3 


ARITHMETIC 


21 


two  vinculmns,  the  last  expression  becomes  24  X  3  -r-  4 
X  2  -r  9  X  5  =  20,  the  same  result  that  would  be  obtained 
by  performing  the  indicated  operations  in  order,  from 
left  to  right. 


EXAMPLES  FOR  PRACTICE 


Find  the  values  of  the  following  expressions: 


(a) 

(8  +  5  —  1)  4-4. 

f(«) 

3 

(b) 

5  X  24  -  32. 

(b) 

88 

(c) 

5  X  24  -r  15. 

(c) 

8 

(d) 

144  -  5  X  24. 

A  ns 

(d) 

24 

(e) 

(1,691  -  540  +  559)  -r 

-  3  X  57. 

(e) 

10 

(O 

2,080  +  120  -  80  X  4 

-  1,670. 

if) 

210 

(g) 

(90  +  60  25)  X  5  - 

29. 

(g) 

1 

W 

90  +  60  25  X  5. 

iw 

1.2 

ARITHMETIC 

(PART  4) 


INVOLUTION 

1.  Involution  is  the  process  of  multiplying  a  number  by 
itself  one  or  more  times.  The  product  obtained  by  this 
process  is  called  a  power  of  that  number. 

Thus,  the  second  power  of  3  is  9,  since  3  X  3  is  9. 

The  third  power  of  3  is  27,  since  3  X  3  X  3  is  27. 

The  fifth  power  of  2  is  32,  since  2x2x2x2X2is  32. 

2.  An  exponent  is  a  small  figure  placed  to  the  right  and 
a  little  above  a  number  to  show  to  what  power  it  is  to  be 
raised,  or  how  many  times  the  number  is  to  be  used  as  a 
factor,  as  the  small  figures  *,  3,  and  5  in  the  following: 

Thus,  3’  =  3  X  3  =  9 

33  =  3x3x3  =  27 

25  =  2x2x2x2x2  =  32 

3.  The  root  of  a  number  is  that  number  which,  used  the 
required  number  of  times  as  a  factor,  produces  the  number. 
In  the  above  cases  3  is  a  root  of  9,  since  3  X  3  is  9.  It  is 
also  a  root  of  27,  since  3  X  3  X  3  is  27.  Also,  2  is  a  root  of 
32,  since  2  X  2  X  2  X  2  X  2  is  32. 

4.  The  second  poiver  of  a  number  is  called  its  square. 
Thus,  5*  is  called  the  square  of  5,  or  5  squared ,  and  its  value 
is  5  X  5  =  25. 

5.  The  third  power  of  a  number  is  called  its  cube.  Thus, 
53  is  called  the  cube  of  5,  or  5  cubed,  and  its  value  is 
5x5x5=  125. 

For  notice  of  copyright ,  see  page  immediately  following  the  title  page 


2 


ARITHMETIC 


§4 


6.  To  find  any  power  of  a  number: 

Example  1. — What  is  the  third  power,  or  cube,  of  35? 

Solution. —  3  5 

3  5 

*  1  7  5 

10  5 

12  2  5 
3  5 

6  12  5 
3  6  7  5 

cube  4  2  8  7  5  Ans. 

Example  2.— What  is  the  fourth  power  of  15? 

Solution. —  1  5 

1  5 

7  5 

1  5 

2  2  5 
1  5 

112  5 

2  2  5 

3  3  7  5 

1  5 

1  6  8  7  5 
3  3  7  5 

fourth  power  5  0  6  2  5  Ans. 


Example  3. —  1.23  —  what? 

Solution. —  1.2 

1.2 
1.4  4 
1.2 

2~88 
14  4 


cube  1.7  2  8  Ans. 


§4 


ARITHMETIC 


3 


Example  4. — What  is  the  third  power,  or  cube,  of  § ? 

Solution. — 

5_33v3_3X3X3_  27 

8  X  8  X  8  8X8X8  512’  AnS‘ 

Rule. — I.  To  raise  a  whole  number  or  a  decimal  to  any 
power ,  use  it  as  a  factor  as  many  times  as  indicated  by  the  exponent. 

II.  To  raise  a  fraction  to  any  power ,  raise  both  the  numer¬ 
ator  and  denominator  to  the  power  indicated  by  the  exponent. 


EXAMPLES  FOR  PRACTICE 
Raise  the  following  to  the  powers  indicated: 

(a)  852 

(b)  (If)’ 

(r)  6.52 

.  (d)  14* 

(e)  (l)3 

(0  (t)3 
Or)  (i)3 

(h)  1.4s 


EVOLUTION 

7.  Evolution  is  the  reverse  of  involution.  It  is  the 
process  of  finding  the  root  of  a  number  that  is  considered 
as  a  power. 

8.  The  square  root  of  a  number  is  that  number  which, 
when  used  twice  as  a  factor,  produces  the  number. 

Thus,  2  is  the  square  root  of  4,  since  2  X  2,  or  2\  =  4. 

9.  The  cube  root  of  a  number  is  that  number  which, 
when  used  three  times  as  a  factor,  produces  the  number. 

Thus,  3  is  the  cube  root  of  27,  since  3  X  3  X  3,  or  33,  =  27. 

10.  The  radical  sign  V,  when  placed  before  a  number, 
indicates  that  some  root  of  that  number  is  to  be  found. 

11.  The  index  of  the  root  is  a  small  figure  placed  over 

and  to  the  left  of  the  radical  sign,  to  show  what  root  is  to  be 
found.  - 


Ans.  • 


(a) 

(b) 

(c) 

(d) 

(e) 

(f) 

(g) 

iw 


7,225 

144 

169 

42.25 

38,416 

2  7 
64 
12  5 
2  16 

3  4  3 
8 

5.37824 


4 


ARITHMETIC 


§4 


Thus,  ^100  denotes  the  square  root  of  100. 

\125  denotes  the  cube  root  of  125. 

^256  denotes  the  fourth  root  of  256,  and  so  on. 
When  the  square  root  is  to  be  extracted,  the  index  is 
generally  omitted.  Thus,  VlOO  indicates  the  square  root 
of  100.  Also,  V225  indicates  the  square  root  of  225. 


SQUARE  ROOT 

12.  The  largest  number  that  can  be  written  with  one  figure 
is  9,  and  9s  =  81;  the  largest  number  that  can  be  written  with 
two  figures  is  99,  and  99“  =  9,801;  with  three  figures  999,  and 
999“  =  998,001;  with  four  figures  9,999,  and  9,999“  =  99,980,- 
001,  etc.  In  each  of  the  above  it  will  be  noticed  that  the 
square  of  the  number  contains  just  twice  as  many  figures  as 
the  number  itself. 

In  order  to  find  the  square  root  of  a  number,  the  first  step 
is  to  find  how  many  figures  there  will  be  in  the  root.  This  is 
done  by  pointing  of!  the  number  into  periods  of  two  figures 
each,  beginning  at  the  right.  The  number  of  periods  will 
indicate  the  number  of  figures  in  the  root.  Thus,  the  square 
root  of  83,740,801  must  contain  4  figures,  since,  pointing 
off 'the  periods,  we  get  83/74'08'01,  or  4  periods;  consequently, 
there  must  be  4  figures  in  the  root.  In  like  manner,  the 
square  root  of  50,625  must  contain  3  figures,  since  we  have 
S'06'25,  3  periods. 

13.  Suppose  that  it  is  desired  to  find  the  square  root 
of  49,729.  Applying  the  method  of  pointing  off  to  the 
number  in  the  example,  gives  4'97'29.  It  will  be  noticed 
that  the  left-hand  period  contains  but  one  figure.  This  must 
necessarily  result  whenever  the  number  contains  an  odd 
number  of  figures. 

Next  draw  a  line  at  the  right  of  the  number  and  find  the 
largest  number  whose  square  is  less  than,  or  equal  to,  4,  the 
first  period.  This  number  is  evidently  2,  since  2x2  =  4. 


4 


ARITHMETIC 


5 


Place  this  number  at  the  right  of  the  line  as  the  first  figure 
of  the  answer.  The  process  up  to  this  point  is  as  follows: 

4'9  7'2  9  (  2 

Subtract  the  square  of  the  first  figure  in  the  answer  from 
the  first  period  and  bring  down  the  next  period,  which  gives 

4'9  7'2  9  (  2 
4 

9  7 

Multiply  the  first  figure  of  the  answer  by  2,  and  use  this 
result  as  a  trial  divisor  to  divide  into  the  remainder  not  con¬ 
sidering  its  last  figure. 

In  this  example,  2x2  =  4,  which,  as  a  trial  divisor,  gives 
2  as  a  result  when  divided  into  9,  which  is  the  remainder 
without  the  last  figure.  Place  this  2  as  the  second  figure  of 
the  answer. 

Place  the  second  figure,  2,  of  the  answer  at  the  right  of  the 
trial  divisor,  making  it  the  unit  figure  of  the  new  divisor, 
which  gives  42  as  the  new  divisor.  This  new  divisor  multi¬ 
plied  by  the  second  figure  of  the  answer  gives  84,  which  is  to 
be  subtracted  from  the  remainder. 

The  process  thus  far  is  as  follows: 

4'9  7'2  9  (  2  2 
4 

4  2)  97 

8  4 

1  3 

Note. — It  frequently  happens  that  the  result  obtained  by  multiplying 
the  complete  trial  divisor  by  the  last  quotient  figure  obtained  by 
dividing  the  remainder,  exclusive  of  its  right-hand  figure,  by  the  first 
trial  divisor  is  larger  than  the  remainder,  in  which  case  the  next 
number  smaller  must  be  tried.  Thus,  in  this  example,  suppose  that 
the  remainder  had  been  80  instead  of  97;  then  the  result  obtained  by 
multiplying  42  by  2  would  have  been  larger  than  the  remainder  and  1 
would  have  been  tried.  The  new  divisor  would  therefore  have  been 
41  and  the  second  figure  of  the  answer  1. 

The  rest  of  the  process  is  simply  a  repetition  of  that 
already  described.  Consequently,  the  next  period  is  now 
brought  down,  which  gives  1,329  as  the  complete  remainder. 


6 


ARITHMETIC 


4 


The  two  figures  in  the  answer  are  multiplied  by  2,  which 
gives  44  as  a  trial  divisor. 

The  trial  divisor  divided  into  the  remainder  without  the 
last  figure,  or  132,  gives  3  as  the  third  figure  of  the  answer. 

Place  the  third  figure,  3,  of  the  answer  at  the  right  of  the 
trial  divisor  used  to  obtain  it,  making  it  the  unit  figure  of 
the  new  divisor,  which  gives  443  as  the  new  divisor;  this 
multiplied  by  3,  gives  1,329,  the  complete  operation  being 
as  follows: 

4'9  7'2  9  (  2  2  3  Ans. 

4 

42  m 

84 

443)1329 
13  2  9 

It  will  be  seen  that  there  is  no  remainder;  223  is  conse¬ 
quently  the  square  root  of  49,729. 

14.  If,  in  any  case,  the  trial  divisor  is  larger  than  the 
remainder  omitting  the  last  figure,  a  cipher  must  be  placed 
in  the  answer  and  also  at  the  right  of  the  trial  divisor,  making 
a  new  trial  d 'visor.  Bring  down  the  next  period  for  a  new 
remainder  and  divide  by  the  trial  divisor  to  obtain  the  next 
figure  of  the  answer. 

Example. — Find  the  square  root  of  255,025. 

Solution. —  2  5'5  0'2  5  (  5 

2  5 

1  0)  5  0 

It  will  be  seen  that  when  the  example  has  been  performed  to  this 
point,  the  trial  divisor  10  will  be  found  to  be  larger  than  5,  which  is 
the  remainder,  omitting  the  last  figure.  Therefore,  place  a  cipher  as 
the  next  figure  of  the  answer;  bring  down  the  next  period  for  a  new 
remainder,  and  multiply  the  figures  in  the  answer  by  2  in  order  to 
obtain  a  trial  divisor.  This  process  is  as  follows: 

2  5'5  0'2  5  (  5  0 
2  5 

1  0  )  5  0 

0  0 

100)  5025 


§4 


ARITHMETIC 


7 


Dividing  the  trial  divisor  100  into  the  remainder  without  the  last 
figure,  or  502,  5  is  obtained.  Place  this  number  as  the  next  figure  of 
the  answer  and  also  as  the  last  figure  of  the  trial  divisor,  which  gives 
1,005  as  the  complete  divisor. 

Multiply  the  complete  divisor  by  the  last  figure  of  the  answer  and 
place  the  result  under  the  remainder.  The  process  is  as  follows: 

2  5' 5  O' 2  5  (  5  0  5  Ans. 

2  5 

10)  50 

0  0 

1005)  5025 

5  0  2  5 

Since  the  product  obtained  by  multiplying  the  complete  divisor  by 
the  last  figure  of  the  answer  exactly  equals  the  remainder,  then  505 
must  be  the  square  root  of  255,025. 

15.  The  square  of  a  decimal  always  contains  twice  as 
many  figures  as  the  number  squared.  For  example,  .1* 
=  .01,  .132  =  .0169,  .7512  =  .564001,  etc.  If  it  be  required 
to  find  the  square  root  of  a  decimal  and  the  decimal  has  not 
an  even  number  of  figures  in  it,  annex  a  cipher.  The  best 
way  to  determine  the  number  of  figures  in  the  root  of  a 
decimal  is  to  begin  at  the  decimal  point,  and,  going  toward 
the  right,  point  off  the  decimal  into  periods  of  two  figures 
each.  Then,  if  the  last  period  contains  but  one  figure, 
annex  a  cipher.  Thus,  the  decimal  .625  is  pointed  off  as 
follows:  .62'50. 

16.  If  the  number  is  a  whole  number  and  decimal,  the 
whole  number  is  pointed  off  to  the  left  and  the  decimal  to 
the  right.  That  is,  commence  at  the  decimal  point  in  both 
cases  and  point  off.  Thus,  the  number  126.423  is  pointed 
off  as  follows:  1'26.42'30. 

17.  The  operation  of  finding  the  square  root  in  all  these 
cases  is  similar  to  that  previously  described,  except  that 
when  the  decimal  point  is  reached,  a  decimal  point  is  placed 
in  the  answer.  This  gives  as  many  decimal  places  in  the 
answer  as  there  are  periods  of  decimals  in  the  number.  An 
example  without  explanation  will  be  given  here. 


8 


ARITHMETIC 


§4 


Find  the  square  root  of  606.6369. 

6'0  6.6  3'6  9  (  2  4.6  3  Ans. 
4 

44)206 
1  7  6 

486)  3063 

2  9  16 

4923)  14769 

1  4  7  6  9 


18.  There  are  comparatively  few  numbers  that  are  exact 
squares  and,  consequently,  it  is  possible  to  find  the  exact 
square  root  in  only  a  small  number  of  cases.  Numbers 
that  have  an  exact  root  are  called  perfect  powers  and  the 
factors  of  the  perfect  powers  are  called  rational  factors. 
Numbers  that  have  no  exact  root  are  called  surds  and  the 
factors  are  called  irrational  factors.  In  the  numbers 
from  1  to  1,000,  inclusive,  there  are  only  42  perfect  powers, 
not  counting  1,  and  of  these  only  30  are  perfect  squares  and 
9  perfect  cubes. 

20  is  an  example  of  a  surd.  This  number  lies  between 
16  (=  42)  and  25  ( =  52);  hence,  the  square  root  of  20,  or 
V20  is  greater  than  4  and  less  than  5  and  is,  therefore,  equal 
to  4  plus  an  interminable  decimal.  In  other  words,  no 
matter  to  how  many  figures  the  square  root  of  20  may  be 
calculated,  the  root  will  never  be  found  exactly. 

Although  the  root  of  a  surd  cannot  be  found  exactly,  as 
close  an  approximate  may  be  obtained  as  is  desired,  since 
the  result  may  be  carried  to  any  required  number  of  decimal 
places  by  annexing  periods  of  two  ciphers  each  to  the  number. 


Example. — 
Solution. — 


§4 


ARITHMETIC 


9 


Example. --What  is  the  square  root  of  3?  Find  the  result  to  five 
decimal  places. 

Solution—  3.0  O'O  O'O  O'O  O'O  0  (  1.7  3  2  0  5  Ans. 

1 

27)200 
1  8  9 

343)  1100 

10  2  9 

3462)  7100 

6  9  2  4 

3464)  17600 

0  0  0  0  0 

346405)  1760000 

1732025 

2  7  9  7  5 


Explanation. — Annexing  five  periods  of  ciphers  to  the 
right  of  the  decimal  point,  the  first  figure  of  the  root  is 
found  to  be  1.  Multiply  1  by  2  in  order  to  obtain  a  trial 
divisor  and  find  how  many  times  it  is  contained  in  the 
remainder  without  the  last  figure,  or  20.  It  is  contained 
10  times,  but  this  is  evidently  too  large.  Trying  9  it  is 
found  that  this  is  also  too  large,  since  29  X  9  =  261  and 
261  is  greater  than  the  remainder  200.  In  the  same  way  it 
is  found  that  8  is  also  too  large.  Trying  7,  7  times  27  is 
189,  a  result  smaller  than  200;  therefore,  7  is  the  second 
figure  of  the  root.  The  next  two  figures,  3  and  2,  are 
easily  found.  The  fifth  figure  in  the  root  is  a  cipher,  since 
the  trial  divisor  3,464  is  greater  than  the  remainder  with¬ 
out  the  last  figure,  or  1,760.  Bringing  down  the  next  period 
and  multiplying  the  number  in  the  answer  by  2  to  obtain 
the  trial  divisor,  it  is  found  that  the  next  figure  in  the 
answer  is  5.  This  gives  the  required  number  of  decimal 
places  in  the  answer. 


Proof. — To  prove  square  root,  square  the  result  obtained. 
If  the  number  is  an  exact  power,  the  square  of  the  root  will 
equal  it;  if  it  is  not  an  exact  power,  the  square  of  the  root, 
plus  the  remainder,  will  equal  it. 


10 


ARITHMETIC 


4 


19.  To  find  the  square  root  of  a  number: 

Rule. — Divide  the  number  into  periods  of  two  figures  each , 
commencing  at  the  right ,  or  if  the  number  contains  a  decimal , 
at  the  decimal  point  and  working  both  ways. 

Find  the  largest  number  that  multiplied  by  itself  will  not  be 
greater  than  the  first  left-hand  period  and  zvrite  this  number  as 
the  first  figure  of  the  answer. 

Square  the  first  figure  of  the  answer  and  subtract  this  square 
from  the  left-hand  period  of  the  number.  To  this  result  annex 
the  next  period  of  the  number  for  a  complete  remainder. 

Multiply  the  first  figure  of  the  answer  by  2  and  use  the  result 
as  a  trial  divisor. 

Divide  the  complete  remainder ,  leaving  off  the  last  figure ,  by 
the  trial  divisor  and  write  the  result ,  or  the  result  diminished , 
as  the  second  figure  of  the  answer. 

To  the  trial  divisor ,  annex  this  second  figure  of  the  answer 
and  use  the  result  as  a  complete  divisor. 

Multiply  the  complete  divisor  by  the  second  figure  of  the  root 
and  subtract  this  result  from  the  complete  remainder. 

To  the  result  thus  obtained  annex  the  next  period  for  a  com¬ 
plete  remainder. 

Proceed  in  this  manner  until  all  the  periods  of  the  number 
have  been  brought  down. 


EXAMPLES  FOR  PRACTICE 


Find  the  square  root  of  the  following: 

(а)  1,936 

(б)  110,889 

(c)  4.9729 

(d)  .0196 

(e)  9,604 

(/)  163,216 


Ans. 


(a)  44 
{b)  333 

( c )  2.23 

(d)  .14 

(e)  98 

( f )  404 


CUBE  ROOT 

20.  Since  cube  and  higher  roots  are  seldom 
required  in  mill  work,  the  student  may,  if  he  so 
desires,  omit  the  following  articles.  No  examples 
relating  to  them  are  included  among  the  Exami¬ 
nation  Questions  at  the  end  of  this  Section; 


4 


ARITHMETIC 


11 


nevertheless,  the  student  is  advised  to  read  care¬ 
fully  the  following  articles,  as  they  contain  much 
that  may  be  of  value  to  him. 

21.  In  the  same  manner  as  in  square  root,  it  can  be 
shown  that  the  periods  into  which  a  number,  whose  cube  root 
is  to  be  extracted,  is  divided  must  contain  three  figures, 
except  that  the  first,  or  left-hand,  period  of  a  whole  or  mixed 
number  may  contain  one,  two,  or  three  figures. 

Suppose  that  it  is  desired  to  find  the  cube  root  of  11,390,625. 
First  separate  the  number  into  periods  of  three  figures  each, 
commencing  at  the  right.  This  gives  H'390'625.  Draw  a 
line  at  the  right  of  the  number  as  was  done  when  extract¬ 
ing  the  square  root  and  find  the  largest  number  whose  cube 
will  not  be  greater  than  the  left-hand  period.  Place  this 
number  as  the  first  figure  of  the  answer. 

In  the  example  given  it  will  be  seen  that  the  left-hand 
period  is  11  and  the  largest  number  whose  cube  will  not  be 
greater  than  11  is  2,  since  33  — ■  27. 

Place  the  2  as  the  first  figure  of  the  answer  and  subtract 
the  cube  of  this  number  from  the  left-hand  period  of  the 
number.  To  the  remainder  thus  obtained  annex  the  next 
period  of  the  number.  The  process  thus  far  is  as  follows: 

1  1'3  9  0'6  2  5  (  2 

8 

3  3  9  0 

It  is  next  necessary  to  find  a  trial  divisor.  This  is  obtained 
by  considering  the  number  in  the  answer  as  tens,  squaring 
it,  and  multiplying  the  result  by  3.  2  considered  as  tens' is  20, 
which  squared  gives  400.  400  multiplied  by  3  gives  1,200  as 

a  trial  divisor. 

Dividing  the  trial  divisor  into  the  remainder  3,390,  it  is 
seen  that  it  is  contained  2  times.  Place  the  2  as  the  second 
figure  of  the  answer.  This  gives  the  following: 

1  1'3  9  0'6  2  5  (  2  2 

8 

20*  X  3  =  1200  |3  3  90 


12  ARITHMETIC  §4 


In  order  to  obtain  a  complete  divisor ,  it  is  necessary  to  add 
together  three  products. 

The  first  product  is  that  obtained  for  the  trial  divisor,  or 
in  this  case  1,200. 

The  second  product  is  obtained  by  considering  the  first 
figure  of  the  answer  as  tens ,  and  multiplying  it  by  the  second 
figure  of  the  answer,  and  by  3.  In  the  example  being  worked, 
the  first  figure  of  the  answer  considered  as  tens  gives  20, 
which  multiplied  by  the  second  figure  of  the  answer,  or  2, 
gives  40,  which  result  multiplied  by  3  gives  120  as  the 
second  product. 

The  third  product  is  obtained  by  squaring  the  second 
figure  of  the  answer,  which  in  this  case  is  22  =  4. 

Adding  these  three  products  together  gives,  for  the  com¬ 
plete  divisor,  1,200  +  120  +  4  =  1,324.  The  example  worked 
to  this  point  is  as  follows: 


1  1'3  9  0'6  2  5  (  2  2 

8 


20*  X  3  =  1  2  0  0 
(20  X  2)  X  3  =  120 

2 2  =  4 


3  39  0 


13  2  4 


Next  multiply  the  divisor  by  the  second  figure  of  the 
answer  and  subtract  the  result  from  the  remainder  previ¬ 
ously  obtained.  To  the  result  annex  the  next  period  for 
a  new  remainder.  The  process  is  as  follows: 


1  1'3  9  0'6  2  5  (  2  2 
8 


202  X  3  =  1  2  0  0 
(20  X  2)  X  3  =  120 

22  =  4 


33  9  0 


13  2  4 


2  6  4  8 


742625 


The  further  processes  are  but  repetitions  of  those  previ¬ 
ously  described  and,  consequently,  will  not  need  much 
explanation. 


§4 


ARITHMETIC 


13 


It  is  next  necessary  to  find  a  trial  divisor  for  the 
remainder.  This  is  done  in  the  same  manner  as  in  the 
first  case,  the  number  in  the  answer  being  considered  as 
tens,  and  squared,  after  which  it  is  multiplied  by  3.  In 
the  example  given  this  will  be  2202  =  48,400,  which,  multi¬ 
plied  by  3,  gives  145,200  for  a  trial  divisor. 

The  trial  divisor  divided  into  the  remainder  gives  5  as 
the  next  figure  of  the  answer. 

Next,  find  the  complete  divisor  by  adding  together  the 
three  products,  taking  the  trial  divisor  as  the  first  product; 
3  times  the  number  already  in  the  answer  considered  as  tens 
multiplied  by  the  next  figure  of  the  answer  for  the  second 
product,  and  the  square  of  the  third  figure  of  the  answer  for 
the  third  product.  Multiply  the  complete  divisor  by  the 
third  figure  of  the  answer  and  subtract  the  result  thus 
obtained  from  the  remainder  previously  obtained.  The 
complete  process  is  as  follows: 


202  X  3  =  12  0  0 

(20  X  2)  X  3  =  120 

T  =  4 

132  4 

2202  X  3  =14  5  2  0  0 
(220  X  5)  X  3  =  3300 

52  = _ 25 

148525 


1  1'3  9  0'6  2  5  (  2  2  5  Ans. 
8 

3  39  0 


26  4  8 
742625 


742625 


Since  the  product  obtained  by  multiplying  the  complete 
divisor  by  the  third  term  of  the  answer  is  just  equal  to  the 
remainder,  the  process  is  complete,  and  the  answer,  or  225, 
is  the  cube  root  of  the  number  11,390,625. 

22.  In  extracting  the  cube  root  of  a  decimal,  proceed  as 
above,  taking  care  that  each  period  contains  three  figures. 
Begin  the  pointing  off  at  the  decimal  point,  going  toward 
the  right.  If  the  last  period  does  not  contain  three  figures, 
annex  ciphers  until  it  does. 


14 


ARITHMETIC 


§  4 


The  number  of  decimal  places  in  the  answer  will  equal 
the  periods  in  the  decimal  whose  root  is  to  be  extracted. 
Example.— What  is  the  cube  root  of  .009129329? 


Solution. —  .0  0  9'1  2  9r3  2  9  (.2  0  9  Ans. 

8 


20*  X  3  = 


12  0  0 


112  9 
0  0  0  0 


200*  X  3  =  1  2  0  0  0  0 
(200  X  9)  X  3  =  5400 

95  =  8  1 


1129329 


125481 


1129329 


Explanation. — In  this  example  it  will  be  noticed  that  in 
the  first  case  the  trial  divisor  1,200  is  greater  than  the 
remainder  1,129;  consequently,  the  next  figure  of  the  answer 
is  0  and  it  is  necessary  to  bring  down  the  next  period  for  a 
new  remainder,  after  which  the  regular  process  is  followed. 

23.  One  example  of  extracting  the  cube  root  of  a  mixed 
number  will  be  given  here  without  any  further  explanation. 

Example.— What  is  the  cube  root  of  47.832147? 


Solution. — 


30a  X  3  =  2  7  0  0 

(30  X  6)  X  3  =  5  4  0 

62  =  3  6 

3  2  7  6 

3602  X  3  =  388800 

(360  X  3)  X  3  =  3240 

32  =  9 


392049 


4  7.8  3  2'1  4  7  (  3.6  3  Ans. 
2  7 

JOSS  2 

1  9  6  5  6 
117  6  1  4  7 


1176147 


Proof. — To  prove  cube  root,  cube  the  result  obtained^  If 
the  given  number  is  an  exact  power,  the  cube  of  the  root 
will  equal  it;  if  not  an  exact  power,  the  cube  of  the  root,  plus 
the  remainder,  will  equal  it. 

24.  To  find  tlie  cube  l'oot  of  a  number: 


Rule.  — Sep  a  ra  te  the  number  into  periods  of  three  figures 
each ,  commencing  at  the  right ,  or  if  the  number  contains  a 
decimal,  at  the  decimal  point  and  working  both  ways. 


§4 


ARITHMETIC 


15 


Find  the  largest  number  whose  cube  is  not  greater  than  the 
left-hand  period  and  write  this  number  as  the  first  figure  of  the 
answer. 

Subtract  the  cube  from  the  left-hand  period  and  to  this  result 
annex  the  next  period  of  the  number. 

Square  the  member  in  the  answer  considered  as  tens ,  and  mul¬ 
tiply  this  result  by  3. 

Use  the  result  thus  obtained  for  a  trial  divisor  to  divide  into 
the  remainder ,  and  place  the  number  resulting  from  this  division , 
or  the  number  diminished ,  as  the  next  figure  of  the  answer. 

To  the  trial  divisor ,  add  the  result  obtained  by  multiplying 
the  first  figure  of  the  answer  considered  as  tens  by  the  second 
figure  of  the  answer  and  by  3;  also  add  the  square  of  -the  second 
figure.  The  sum  thus  obtained  is  the  complete  divisor. 

Multiply  the  complete  divisor  by  the  second  figure  of  the 
answer  and  subtract  the  result  thus  obtained  from  the  remainder. 

To  this  result  annex  the  next  period  of  the  number  and  proceed 
in  the  manner  described  until  all  the  periods  have  been  tised. 


EXAMPLES  FOR  PRACTICE 
Find  the  cube  root  of  the  following  numbers: 


(«) 

1,860,867 

» 

123 

(0 

185,193 

(b) 

57 

(0 

636,056 

Ans.  • 

{c) 

86 

(d) 

87,528.384 

id) 

44.4 

(e) 

74,088 

{e) 

42 

(f) 

257,259,456 

1(0 

636 

ROOTS 

OF  FRACTIONS 

25.  •If  the  given  number  is  in  the  form  of  a  fraction,  and 
it  is  required  to  find  some  root  of  it,  the  simplest  and  most 
exact  method  is  to  reduce  the  fraction  to  a  decimal  and  extract 
the  required  root  of  the  decimal.  If,  however,  the  numerator 
and  denominator  of  the  fraction  are  perfect  powers,  extract 
the  required  root  of  each  separately,  and  write  the  root  of 
the  numerator  for  a  new  numerator,  and  the  root  of  the 
denominator  for  a  new  denominator. 


16 


ARITHMETIC 


§4 


Example 
Solution.— 


1. — What  is  the  square  root 

IT  =  =  3 

\64  _  yl64  8' 


of  A? 

Ans. 


Example  2. — What  is  the  square  root  of  f  ? 

Solution. — Since  f  =  .625,  Vf  =  V. 625  =  .7906.  Ans. 
Example  3. — What  is  the  cube  root  of  ff  ? 


Solution. — 


Example  4. — What  is  the  cube  root  of  i? 


Solution. — Since  \  =  .25 =  C25  =  .62996. 


Ans. 


Rule. — Extract  the  required  root  of  the  numerator  and 
denominator  separately ;  or,  reduce  the  fraction  to  a  decimal, 
and  extract  the  root  of  the  decimal. 


TABLE  METHOD  OF  EXTRACTING  SQUARE 
AND  CUBE  ROOTS 

26.  In  any  number,  the  figures  beginning  with  the  first 
digit*  at  the  left  and  ending  with  the  last  digit  at  the  right, 
are  called  the  significant  figures  of  the  number.  Thus, 
the  number  405,800  has  the  four  significant  figures  4,  0,  5,  8; 
and  the  number  .000090067  has  the  five  significant  figures 
9,  0,  0,  6,  and  7. 

The  part  of  a  number  consisting  of  its  significant  figures 
is  called  the  significant  pai't  of  the  number.  Thus,  in  the 
number  28,070,  the  significant  part  is  2807;  in  the  number 
.00812,  the  significant  part  is  812;  and  in  the  number  170.3, 
the  significant  part  is  1703. 

In  speaking  of  the  significant  figures  or  of  the  significant 
part  of  a  number,  we  consider  the  figures,  in  their  proper 
order,  from  the  first  digit  at  the  left  to  the  last  digit  at  the 
right,  but  we  pay  no  attention  to  the  position  of  the  decimal 
point.  Hence,  all  numbers  that  differ  only  in  the  position  ot 
the  decimal  point  have  the  same  significant  part.  For  example, 
.002103,  21.03,  21,030,  and  210,300  have  the  same  significant 
figures  2,  1,0,  and  3,  and  the. same  significant  part  2103. 


*A  cipher  is  not  a  digit. 


§4 


ARITHMETIC 


17 


SQUARES  AND  CUBES 


No. 

Square 

Cube 

1.0 

I  .OO 

1. 000 

1 .1 

1 .21 

I-33I 

1.2 

1.44 

1.728 

1.3 

1.69 

2.197 

i-4 

1.96 

2-744 

i-5 

2.25 

3-375 

1.6 

2.56 

4.096 

i-7 

2.89 

4-913 

i.8 

3-24 

5-832 

1-9 

3.61 

6.859 

2.0 

4.00 

8.000 

2.1 

4.41 

9.261 

2.2 

4.84 

10.648 

2-3 

5-29 

12.167 

2.4 

5-76 

13.824 

2-5 

6.25 

i5-625 

2.6 

6.76 

I7.576 

2.7 

7.29 

19.683 

2.8 

7.84 

21.952 

2.9 

8.41 

24.389 

3-0 

9.00 

27.000 

3-i 

9,6 1 

29.791 

3-2 

10.24 

32.768 

3-3 

10.89 

35-937 

3-4 

11.56 

39-304 

3-5 

12.25 

42.875 

3-6 

12.96 

46.656 

3-7 

13.69 

50.653 

3-8 

14.44 

54-872 

3-9 

15.21 

59-319 

4.0 

16.00 

64.000 

4.1 

16.81 

68.921 

4.2 

17.64 

74.088 

4-3 

18.49 

79.507 

4-4 

19.36 

85.184 

4-5 

20.25 

91.125 

4.6 

21 . 16 

97-336 

4-7 

22.09 

103.823 

4-8 

23.04 

no. 592 

4-9 

24.01 

117.649 

5-0 

25.00 

125.000 

5-i 

26.01 

132.651 

5-2 

27.04 

140.608 

5-3 

28.09 

148.877 

5-4 

29.16 

157.464 

No. 

Square 

Cube 

5-5 

30.25 

166.375 

5-6 

3I.36 

175.616 

5-7 

32.49 

185.193 

5-8 

33-64 

I95-II2 

5-9 

34-8i 

205.379 

6.0 

36.00 

2x6.000 

6.1 

37.21 

226.981 

6.2 

38.44 

238.328 

6.3 

39.69 

250.047 

6.4 

40.96 

262.144 

6.5 

42.25 

274.625 

6.6 

43-56 

287.496 

6-7 

44.89 

300.763 

6.8 

46.24 

314.432 

6.9 

47.6i 

328.509 

7-0 

49.00 

343.000 

7-i 

50.41 

357.911 

7.2 

51.84 

373.248 

7-3 

53-29 

389.017 

7-4 

54-76 

405.224 

7-5 

56.25 

421.875 

7.6 

57-76 

438.976 

7-7 

59-29 

456.533 

7-8 

60.84 

474.552 

7-9 

62.41 

493.039 

8.0 

64.00 

512.000 

8.1 

65.61 

531.441 

8.2 

67.24 

551.368 

8-3 

68.89 

571.787 

8.4 

70.56 

592.704 

8-5 

72.25 

614.125 

8.6 

73.96 

636.056 

8-7 

75.69 

658.503 

8.8 

77-44 

681.472 

8.9 

79.21 

704.969 

9.0 

81.00 

729.000 

9.1 

82.81 

753.571 

9.2 

84.64 

778.688 

9-3 

86.49 

804.357 

9.4 

88.36 

830.584 

9-5 

90.25 

857.375 

9.6 

92.16 

884.736 

9-7 

94.09 

912.673 

9.8 

96.04 

941 . 192 

9.9 

98.01 

970.299 

18 


ARITHMETIC 


4 


27.  Figuring’  the  root  of  a  whole  number  to  one  decimal 
place  is  often  accurate  enough  for  practical  purposes  in  mill 
work  and  the  following  method  of  obtaining  the  square  and 
cube  roots  of  numbers,  which  will  in  almost  every  case  give 
sufficiently  accurate  results,  will  be  found  to  be  of  advantage. 

28.  By  means  of  the  table  which  contains  the  squares  and 
cubes  of  numbers  from  1  to  10,  varying  by  tenths,  the  first 
three,  and  frequently  the  first  four,  significant  figures  of  the 
square  root  or  cube  root  of  any  number  can  be  readily 
determined. 

29.  By  the  aid  of  this  table,  the  first  two  significant 
figures  of  the  root  can  be  obtained  directly  and  one  more  by 
a  slight  calculation.  For  example,  suppose  that  it  is  desired 
to  find  the  first  three  significant  figures  of  V5,269.73.  Point¬ 
ing  off  into  periods  and  moving  the  decimal  point  so  that 
it  falls  between  the  first  and  second  periods,  the  number 
becomes  52.69/73;  in  other  words,  the  significant  figures  of 
V5,269.73  are  the  same  as  of  V5 2.6973.  Since  four  figures 
only  are  given  in  the  table,  reduce  the  given  nifmber  to  four 
figures.  The  problem  then  becomes:  find  the  first  three  fig¬ 
ures  of  V52.70.  Referring  to  the  table,  52.70  lies  between 
51.84  (  =  7.22)  and  53.29  (  =  7.32);  hence,  the  first  two  figures 
of  the  root  are  7.2.  Find  the  difference  between  the  two 
numbers  in  the  table  between  which  the  given  number  falls 
and  call  it  the  first  difference;  thus,  53.29  —  51.84  =  1.45  = 
the  first  difference.  Find  the  difference  between  the  smaller 
number  in  the  table  and  the  given  number  and  call  it  the 
second  difference;  thus,  52.70  —  51.84  =  .86  =  the  second 
difference.  Divide  the  second  difference  by  the  first  differ¬ 
ence,  and  the  first  figure  of  the  quotient,  if  the  quotient 
is  .05  or  greater,  will  be  the  third  figure  of  the  root,  when 
reduced  to  one  figure.  If  the  quotient  is  less  than  .05,  the 
third  figure  of  the  root  is  a  cipher.  Thus,  .86  -r-  1.45  =>  .59, 
or  .6  when  reduced  to  one  figure.  Therefore,  the  first  three 
figures  of  V52.70  are  7.26.  Since  the  integral  part  of  the 
given  number  contains  two  periods,  there  are  two  figures  in 


§4 


ARITHMETIC 


19 


the  integral  part  of  the  root;  therefore,  V5.269.73  =  72.6  to 
three  figures. 

30.  The  cube  root  is  found  to  three  significant  figures  in 
exactly  the  same  way,  as  shown  in  the  following  example: 

Example. — Find  the  first  three  figures  of  V~0625. 

Solution. — Pointing  off  and  placing  the  decimal  point  between  the 
first  and  second  significant  periods,  the  result  is  62.500.  Referring  to 
the  table,  the  first  two  figures  of  the  root  are  3.9;  the  first  difference  is 
64.000  -  59.319  =  4.681;  the  second  difference  is  62.500  -  59.319 
=  3.181;  3.181  -r-  4.681  =  .67,  or  .7  to  one  figure.  Therefore,  V62.5 
=  3.97,  and  V.0625  =  .397  to  three  significant  figures.  Ans. 


TO  EXTRACT  OTHER  ROOTS  THAN  THE  SQUARE 

AND  CUBE  ROOTS 

31.  By  two  or  more  operations  certain  roots  other  than 
square  and  cube  roots  may  be  extracted;  thus,  to  find  the 
fourth  root  of  a  number  it  is  only  necessary  to  find  the 
square  root  of  the  number  and  then  extract  the  square  root 
of  the  result  thus  obtained,  or  to  find  the  sixth  root  first 
extract  the  cube  root  and  then  the  square  root,  etc. 

Example  1. — What  is  the  fourth  root  of  256? 

Solution. —  V256  =  16;  Vl6  =  4 

Therefore,  V256  =  4.  Ans. 

In  this  example,  V256,  the  index  is  4,  which  equals  2X2.  The  root 
indicated  by  2  is  the  square  root;  therefore,  the  square  root  is  extracted 
twice. 

Example  2. — What  is  the  sixth  root  of  64? 

Solution. —  >/64  =4;  Vi  =  2 

Therefore,  V64  =  2.  Ans. 

In  this  example,  V64,  the  index  is  6,  which  equals  3  X  2.  The  root 
indicated  by  3  is  the  cube  root;  therefore,  the  cube  and  square  roots 
are  extracted  in  succession. 

Example  3. — What  is  the  sixth  root  of  92,873,580  to  two  decimal 
places? 

Solution. —  6  =  3X2.  Hence,  extract  the  cube  root,  and  then 
extract  the  square  root  of  the  result.  V92,873,580  =  4.52.8601,  and 
V452.8601  =  21.28.  Ans. 


20  ARITHMETIC  §4 

It  matters  not  which  root  is  extracted  first,  but  it  is  probably  easier 
and  more  exact  to  extract  the  cube  root  first. 

Rule. — Separate  the  index  of  the  required  root  into  its  factors 
(3's  and  2’s),  and  extract ,  successively ,  the  roots  indicated  by  the 
several  factors  obtained.  The  fuial  result  will  be  the  required 
root. 


ARITHMETIC 

(PART  5) 


DENOMINATE  NUMBERS 

1.  A  simple  number  expresses  units,  either  abstract  or 
of  a  single  denomination;  thus,  2  dollars  or  7  gallons  are 
simple  numbers. 

2.  A  compound  number  is  a  collection  of  concrete 
units  of  several  denominations;  thus,  10  pounds  and  4  ounces 
or  5  gallons  and  2  quarts  are  compound  numbers. 

3.  A  denominate  number  is  a  concrete  number  that 
may  be  changed  to  a  different  denomination  without  chan¬ 
ging  its  value;  thus,  2  gallons  equal  8  quarts,  both  numbers 
expressing  the  same  quantity. 

4.  Reduction,  as  applied  to  arithmetic,  is  the  changing 
or  reducing  of  a  number  into  one  of  a  different  denomination 
but  having  an  equal  value. 

5.  Reduction  descending;  is  the  process  of  changing  a 
number  of  one  denomination  into  a  number  of  a  lower 
denomination,  as,  for  instance,  gallons  to  quarts  or  dollars  to 
cents.  Reduction  descending  involves  the  process  of  multi¬ 
plication. 

6.  Reduction  ascending  is  the  process  of  changing  a 
number  of  one  denomination  into  a  number  of  a  higher 
denomination,  as,  for  instance,  quarts  to  gallons  or  cents  to 
dollars.  Reduction  ascending  is  the  reverse  of  reduction 
descending  and  involves  the  process  of  division. 

For  notice  of  copyright,  see  page  immediately  following  the  title  page 

i  5 


2 


ARITHMETIC 


§5 


MEASURES 

7.  A  measure  is  a  standard  unit ,  established  by  law  or 
custom,  by  which  quantity  of  any  kind  is  measured. 


MEASURE  OF  MONEY 
UNITED  STATES  MONEY 

8.  The  following  table  is  the  one  used  for  the  currency 


of 

the  United  States 

• 

Table 

10 

mills  (ra.) . 

.  .  =  1 

cent . 

.  ct 

10 

cents . 

,  .  .  =  1 

dime . 

.  d. 

10 

dimes . 

,  .  .  =  1 

dollar . 

.  $ 

10 

dollars . 

.  .  .  =  1  eagle . 

.  E 

Mills 

Cents 

Dimes 

Dollars  Eagle 

10  = 

1 

100  = 

10 

-  1 

1,000  = 

100 

=  10 

=  1 

10,000  = 

1,000 

=  100 

=  10  =  1 

9.  The  various  denominations  of  United  States  currency 
are  based  on  a  decimal  system,  the  unit  being  1  dollar;  thus, 
one-tenth  of  1  dollar  is  1  dime  and  ten  times  1  dollar  is  1  eagle. 

Dollars  are  separated  from  cents  and  mills  by  a  decimal 
point,  the  cents  occupying  the  first  two,  and  the  mills,  the 
third  place  to  the  right  of  the  point,  since  cents  represent 
hundredth  parts  of  a  dollar  and  mills,  thousandth  parts;  thus, 
$25,487  is  read  twenty-five  dollars  forty-eight  cents  and  seven 
mills. 

When  the  number  of  cents  in  an  expression  of  dollars  and 
decimal  parts  of  a  dollar  is  less  than  ten,  a  cipher  is  inserted 
between  the  decimal  point  and  the  figure  denoting  the  num¬ 
ber  of  cents,  since  cents  represent  hundredth  parts  of  a 
dollar,  thus  $14.06. 


§5 


ARITHMETIC 


3 


MEASURES  OF  WEIGHT 
AVOIRDUPOIS  WEIGHT 

10.  Avoirdupois  weight  is  commonly  used  for  weigh¬ 
ing  goods  of  all  descriptions,  except  in  compounding 
medicines,  and  for  all  metals  except  gold,  silver,  etc. 


.  Table 

16  drams  (dr.) . =  1  ounce  . oz. 

16  ounces . =  1  pound . lb. 

100  pounds . =  1  hundredweight  .  .  .  cwt. 

20  hundredweight . =  1  ton . T. 

Drams  Ounces  Pounds  Ton 

W  r.KiH  1 

16  =  1 

256  =  16  1 

25,600  -  1,600  =  100  =  1 

512,000  =  32,000  =  2,000  =  20  =  1 


11.  A  long  ton  is  equal  to  2,240  pounds  and  is  used 
in  connection  with  large  lots  of  merchandise,  notably  iron 
and  coal,  when  bought  and  sold  by  the  wholesale.  A 
long  hundredweight  is  112  pounds.  The  long  ton  and 
long  hundredweight  are  used  in  the  United  States  Custom 
Houses.  Unless  otherwise  stated,  the  short  ton  (2,000 
pounds)  and  short  hundredweight  (100  pounds)  are  always 
referred  to. 

12.  An  adaptation  of  the  avoirdupois  weight  that  is 
used  in  mill  work  for  weighing  yarn,  roving,  etc.,  is  as 
follows: 

Table 

27.34+  grains  =  1  dram 

437.50  grains  =  16  drams  =  1  ounce 

7,000.00  grains  =  256  drams  =  16  ounces  =  1  pound 

TROY  WEIGHT 

13.  Troy  weight  is  used  in  weighing  gold  and  silver 
and  in  mints  and  jewelers’  shops  generally. 

Table 


24  grains  (gr.) 
20  pennyweights 
12  ounces  .  .  . 


1  pennyweight  ....  pwt. 

1  ounce . oz. 

1  pound . lb. 


4 


ARITHMETIC 


§5 


Grains 

Penny¬ 

weights 

Ounces 

Pound 

24 

480 

=  1 

=  20  = 

=  1 

5,760 

=  240  = 

=  12 

=  1 

APOTHECARIES’  WEIGHT 


14.  Apothecaries’  weight  is  used  in  mixing  medicines. 


Table 

20  grains  (gr.)  =  1  scruple  .... 

3  scruples . =  1  dram . 

8  drams . =  1  ounce . 

12  ounces . =  1  pound  .... 

Grains  Scruples  Drams  Ounces  Pound 

20  =  1 

60  =  3  =  1 

480  =  24  =  8  =  1 

5,760  =  288  -  96  =  12  =  1 


sc. 

dr. 


.  oz. 
.  lb. 


MEASURES  OF  QUANTITY 

LIQUID  MEASURE 

15.  Liquid  measure  is  commonly  employed  for  meas¬ 
uring  the  more  common  liquids. 


4  gills  (gi.)  .  . 

2  pints  .  .  .  . 

4  quarts  .  .  . 

31|  gallons  .  .  . 

63  gallons  .  .  . 

Gills  Pints 
4  =  1 

8  =  2 
32  =  8 

1,008  =  252 

2,016  =  504 


Table 

.  .  =  1  pint . pt. 

.  .  =  1  quart . qt. 

.  .  =  1  gallon . gal. 

.  .  =  1  barrel . bbl. 


=  1  hogshead . hhd. 


Quarts 

1 

4 

Gallons 

=  1 

Barrels 

Hogshead 

126 

=  311 

1 

252 

=  63 

2  = 

1 

APOTHECARIES’  FLUID  MEASURE 

16.  Apothecaries’  fluid  measure  is  used  for  measuring 
medicines  and  other  liquids  that  are  used  in  small  quantities. 


Table 

60  minims  (min.) . 

8  fluid  drachms  : . 

16  fluid  ounces  . 

8  pints . 


1  fluid  drachm 
1  fluid  ounce 
1  pint 
1  gallon 


§5 


ARITHMETIC 


5 


DRY  MEASURE 

17.  Dry  measure  is  used  for  measuring  grains,  fruits, 
vegetables,  salt,  coal,  and  similar  substances. 


Table 

2  pints  (pt.) 

= 

1  quart . 

.  qt. 

8  quarts  .  . 

= 

1  peck  . 

.  pk. 

4  pecks  .  .  . 

1  bushel  . 

bu. 

8  bushels  .  . 

= 

1  quarter . 

.  qr. 

Pints 

Quarts 

Pecks 

Bushels  Quarter 

2 

=  1 

16 

=  8 

=  1 

64 

=  32 

=  4 

=  1 

512 

=  256 

=  32 

=  8  =  1 

MEASURES  OF  LENGTH 

LINEAR,  OR  LONG,  MEASURE 

18.  Linear,  or  long,  measure  is  used  for  measuring 
distances  of  any  magnitude  and  in  any  direction. 


12  inches  (in.)  or  (")  .  .  . 

Table 

1  foot  .  .  . 

...  ft.  or  (') 

3  feet  . 

= 

1  yard  .  .  . 

.  .  .  yd. 

51  yards,  or  161  feet  .... 

= 

1  rod  .  .  . 

.  .  .  rd. 

40  rods  . 

= 

1  furlong 

.  .  .  fur. 

8  furlongs,  or  320  rods  .  . 

= 

1  mile  .  .  . 

.  .  .  m. 

Inches  Feet 

Yards 

Rods  Furlongs  Mile 

12  =  1 

36  =  3  = 

198  =  161  = 

1 

51 

1 

7,920  =  660  = 

220 

= 

40  = 

1 

63,360  =  5,280  = 

1,760 

= 

320  = 

8  =  1 

SURVEYORS’ 

MEASURE 

19.  Surveyors’  measure 

is 

used  by 

surveyors  in 

measuring  land,  roads,  etc. 

Table 

7.92  inches . = 

1  link  .  .  . 

. I. 

25  links  . 

= 

1  pole  .  .  . 

. P- 

100  links,  4  poles,  or  66  feet  . 

= 

1  chain  .  .  . 

. cha. 

10  chains . 

= 

1  furlong  .  . 

. fur. 

8  furlongs,  or  80  chains  .  . 

= 

1  mile  .  .  . 

. m. 

6 


ARITHMETIC 


5 


Gunter’s  cliain  is  66  feet  in  length  and  divided  into  100 
links  and  is  used  in  ordinary  land  surveys,  but  for  locating 
roads  and  laying  out  public  works  an  engineer’s  chain  100' 
feet  in  length  and  containing  100  links  is  used. 


MEASURES  OF  AREA 

SQUARE  MEASURE 

20.  Square,  or  surface,  measure  is  used  in  measuring 
surfaces  and  areas  of  all  kinds. 

Note. — A  square  is  a  figure  having  four  equal  sides  and  all  of  its 
angles  equal:  a  square  inch ,  therefore,  is  the  amount  of  surface 
enclosed  by  a  square  the  sides  of  which  are  1  inch  long;  similarly  a 
square  foot  or  square  mile  is  the  amount  of  surface  or  the  area  enclosed 
in  squares  the  sides  of  which  are  each  1  foot  or  1  mile  long. 

Table 

144  square  inches  (sq.  in.)  .  .  .  =  1  square  foot . sq.  ft. 


9  square  feet . =  1  square  yard . sq.  yd. 

301  square  yards,  or  1  ,  ,  , 

,  ^  >  . =  1  square  rod . sq.  rd. 

2721  square  feet  j 

160  square  rods . =  1  acre . A. 

640  acres . =  1  square  mile . sq.  m. 

Square  Square  Square  Square  ,  Square 

Inches  Feet  Yards  Rods  Mile 

144  =  1 

1,296  =  9  =  1 

39,204  =  2721  =  301  =  1 

6,272,640  =  43,560  =  4,840  =  160  =  1 


4,014,489,600  =  27,878,400  =  3,097,600  =  102,400  =  640  =  1 


MEASURES  OF  SOLIDITY  OR  VOLUME 

CUBIC  MEASURE 

21.  Cubic,  or  solid,  measure  is  used  for  measuring 
such  materials  as  have  length,  breadth,  and  thickness,  as 
timber,  stone,  and  other  similar  materials. 

Note. — A  cube  is  a  body  bounded  by  six  square  and  equal  sides; 
therefore,  a  cubic  inch  is  the  volume  contained  in  a  cube  having  six 
faces  each  1  inch  square,  and  similarly  a  cubic  foot  or  a  cubic  yard  is 
the  volume  of  cubes  with  sides  1  foot  or  1  yard  square. 


§5  ARITHMETIC  7 

Table 

1,728  cubic  inches  (cu.  in.)  .  .  .  =  1  cubic  foot . cu.  ft. 

27  cubic  feet . =  1  cubic  yard . cu.  yd. 

Cubic  Inches  Cubic  Feet  Cubic  Yard 

1,728  =  1 

46,656  =  27  =  1 

22.  A  table  that  is  used  for  measuring  wood  for  fuel  is 
as  follows: 

Table 

16  cubic  feet . =  1  cord  foot  . c.  ft. 

8  cord  feet,  or 1  .  , 

> . =  1  cord  . c. 

128  cubic  feet  j 


MEASURE  OF  TIME 
23.  Time  is  measured  as  follows: 


Table 


60  seconds  (sec.) 
60  minutes  .  . 


1  minute . min. 

1  hour . hr. 


24  hours  .  .  . 

7  days  .... 
365i  days,  or 
52  weeks  H  days 


1  day . d. 

1  week . wk. 

1  year . yr. 


Seconds 

Minutes 

Hours 

Days 

Weeks 

60 

=  1 

3,600 

60 

=  1 

86,400 

=  1,440 

=  24 

=  1 

604,800 

=  10,080 

=  168 

=  7 

=  1 

31 ,557,600 

=  525,960 

=  8,766 

=  365-1 

=  52^ 

Note.— For  convenience  it  is  customary  to  reckon  365  days  as  a  year  and  call 
every  fourth  year  366  days,  placing  the  extra  day  in  the  month  of  February,  which 
then  has  29  days.  This  is  known  as  a  leap  year.  A  year  is  equal  to  12  months 
(mo.)  and  for  convenience  a  month  is  considered  as  30  days. 


ANGULAlt  MEASURE 

24.  Angular,  or  circular,  measure  is  employed  for 
measuring  the  angle  between  two  lines  or  surfaces  and  also 
for  measuring  circles,  latitude,  longitude,  etc. 

Table 


60  seconds  (") . =  1  minute .  ' 

60  minutes  . =  1  degree . ° 

90  degrees . =  1  right  angle,  or  quadrant  L 

360  degrees,  or  4  L  . =  1  circumference  ....  cir. 


8 


ARITHMETIC 


§5 


Seconds 

Minutes 

Degrees 

Quadrants  Circumference 

60 

=  1 

3,600 

=  60 

=  1 

324,000 

-  5,400 

=  90 

=  1 

1,296,000 

=  21,600 

=  360 

=  4  1 

MISCELLANEOUS  MEASURES 


1  pound  sterling  (^) . 

1  league  (lea.) . 

1  fathom  . 

1  knot,  or  nautical  mile . 

1  meter . 

1  decimeter . 

1  centimeter . 

1  millimeter . 

1  dozen  (doz.) . 

1  gross  . 

1  great  gross  . 

1  quire  . 

1  ream  . 

1  large  ream . 

1  perch . 

1  gallon . 

1  tierce . 

1  puncheon  . 

1  carat  . 

1  butt . 

1  bushel . 

1  palm . 

1  hand  . 

1  span  . 

1  gallon  of  water  (U.  S.  Standard) 
1  gallon  of  water  (British  Imperial 

gallon)  . 

1  cubic  foot . 


=  $4.8665 
=  3  miles 
=  6  feet 
-  1H  miles 
=  39.37  inches 
=  3.937  inches 
=  .3937  inch 
=  .03937  inch 
=  12  articles 
=  12  dozen 
=  12  gross 
=  24  sheets  of  paper 
=  20  quires 
=  500  sheets 
=  24f  cubic  feet 
=  231  cubic  inches 
—  42  gallons 
=  2  tierces 
=  3|  grains  (troy) 

=  108  gallons 
=  2,150.42  cubic  inches 
=  3  inches 
=  4  inches 
=  9  inches 

=  231  cubic  inches  =  8.355  pounds 

=  277  cubic  inches  =  10  pounds 
=  7.481  gallons 


§5 


ARITHMETIC 


9 


REDUCTION  OF  COMPOUND  QUANTITIES 

25.  Reducing  a  quantity  to  a  lower  denomination,  or 
reduction  descending: 


Example  1. — Reduce  1  T.,  3  cwt.,  45  lb.  and  11  oz.  to  ounces. 


Solution. — 


1  T.  3  cwt.  45  lb.  11  oz. 

20  cwt.  in  1  T. 

2  0  cwt. 

3  cwt. 


2  3  cwt. 

10  0  lb.  in  1  cwt. 
2  3  0  0  lb. 

_ 4  5  lb. 

2  3  4  5  lb. 

1  6  oz.  in  1  lb. 


1  4  0  7  0 

2  3  4  5 

3  7  5  2  0  oz. 

1  1  oz. 


3  7  5  3  1  oz.  Ans. 


Explanation. — One  ton  multiplied  by  the  number  of 
hundredweight  in  1  ton  (20)  =  20  hundredweight.  20  hun¬ 
dredweight  plus  3  hundredweight  =  23  hundredweight, 
which  when  multiplied  by  the  number  of  pounds  in  one 
hundredweight  (100)  =  2,300  pounds;  adding  45  pounds 
=  2,345  pounds.  This  product  multiplied  by  the  number 
of  ounces  in  1  pound  (16)  =  37,520  ounces;  adding  11  ounces 
=  37,531  ounces,  the  number  of  ounces  in  1  ton  3  hun¬ 
dredweight  45  pounds  11  ounces. 

Example  2. — Reduce  4  m.  25  rd.  10  ft.  to  ft. 

Solution. —  4  m.  25  rd.  10  ft. 

3  2  0  rd.  in  1  m. 

1  2  8  0  rd. 

2  5  rd. 

13  0  5 

1  6  i  ft.  in  1  rd. 

6  5  2  4 
7  8  3  0 
13  0  5 

2  1  5  3  2  4  ft. 

1  0  ft. 


2  1  5  4  2  4  ft.  Ans. 


10 


ARITHMETIC 


§5 


26.  To  reduce  a  compound  quantity  to  a  lower 
denomination: 

Rule. — Multiply  the  given  number  of  units  of  the  highest 
denomination  by  the  member  of  emits  of  the  next  lower  denomi¬ 
nation  required  to  make  a  unit  of  the  highest  denomination ,  and 
to  the  product  add  the  corresponding  denomination  of  the  multi¬ 
plicand  if  there  be  any.  Proceed  in  like  manner  until  the  required 
denomination  is  reached. 


EXAMPLES  FOR  PRACTICE 


Reduce: 

(а)  2  lb.  (avoirdupois)  to  ounces. 

(б)  7  T.  17  cwt.  48  lb.  8  oz.  14  dr.  (avoir¬ 

dupois)  to  drams. 

(c)  1  lb.  11  oz.  to  grains.  (See  Art.  12.) 

( d )  31b.  6  oz.  to  grains.  (See  Art.  12.) 


[(a) 

(b) 


Ans.  • 


32  oz. 

4,031,630  dr. 


(c)  11,812.5  gr. 
{d)  23,625  gr. 


27.  Reducing-  a  compound  quantity  to  a  higher  denomina¬ 
tion,  or  reduction  ascending: 

Example  1. — Reduce  35,678  drams  to  higher  denominations. 
Solution. —  1  6  ) 3  5  6  7  8  dr. 

1  6)  2  2  2  9  oz.  and  14  dr. 

1  0  0  )  1  3  9  lb.  and  5  oz. 

1  cwt.  and  39  lb. 

1  cwt.  39  lb.  5  oz.  14  dr.  Ans. 


Explanation. — Dividing  the  total  number  of  drams  by 
the  drams  in  1  ounce  gives  2,229  ounces  and  14  drams  left 
over.  Dividing  the  ounces  by  the  ounces  in  1  pound, 
139  pounds  is  obtained  and  5  ounces  left  over;  dividing 
the  pounds  by  the  pounds  in  1  hundredweight,  1  hundred¬ 
weight  and  39  pounds  is  obtained.  Thus,  35,678  drams 
equals  1  hundredweight  39  pounds  5  ounces  14  drams. 

Example  2. — Reduce  454,621  pt.  (dry  measure)  to  highest  denomina¬ 
tions. 

2  HA  4  HA 

8  )2  2  7  3  1  0  qt.  and  1  pt. 

4  )2  8  4  1  3  pk.  and  6  qt. 

8)7103  bu.  and  1  pk. 

8  8  7  qr.  and  7  bu. 

887  qr.  7  bu.  1  pk.  6  qt.  1  pt.  Ans. 


Solution. — 


§5 


ARITHMETIC 


11 


28.  To  reduce  a  quantity  to  a  higher  denomina¬ 
tion: 

Rule. — Divide  by  the  number  of  units  required  to  make  one 
unit  of  the  next  higher  denomination. 

Divide  the  quotient  and  each  successive  quotient  thus  acquired 
in  like  manner  until  the  required  denomination  is  reached. 

The  last  quotient  and  the  sevetal  remainders  arranged  in  the 
order  of  their  successive  denominations  is  the  answer  required. 


EXAMPLES  FOR  PRACTICE 


Reduce  the  following  numbers  to  highest  denominations: 


( a )  2,789  oz.  (avoirdupois). 

(b)  348  dr.  (avoirdupois). 

( c )  16,001  oz.  (avoirdupois) . 

(d)  32,645  dr.  (avoirdupois). 


Ans. 


'  (a)  1  cwt.  74  lb.  5  oz. 

(b)  1  lb.  5  oz.  12  dr. 

(c)  10  cwt.  1  oz. 

.(d)  1  cwt.  27  lb.  8  oz.  5  dr. 


REDUCTION  OE  UNITED  STATES  MONEY 

29.  Since  United  States  currency  varies  as  a  decimal 
system,  the  following  rules  will  enable  the  student  to  reduce 
it  to  a  lower  denomination: 

Rule. — I.  To  reduce  dollars  to  cents,  annex  two  ciphers. 

II.  To  reduce  dollars  to  mills,  annex  three  ciphers. 

III.  To  reduce  cents  to  mills,  annex  one  cipher. 

I V .  Dollars ,  cents,  and  mills  expressed  as  a  single  number 
arc  reduced  to  mills  by  removing  the  decimal  point. 

V.  Dollars  and  cents  are  reduced  to  mills  by  removing  the 
point  and  annexing  one  cipher. 

30.  To  reduce  United  States  currency  to  a  higher 
denomination: 

Rule. — I.  To  reduce  mills  to  cents,  divide  by  ten  or  point  off 
with  a  decimal  point  one  place  on  the  right. 

II.  To  reduce  cents  to  dollars,  divide  by  one  hmidred  or  point 
off  two  places. 

III.  To  reduce  mills  to  dollars,  point  off  three  places  or 
divide  by  one  thousand. 


12 


ARITHMETIC 


5 


ADDITION  OF  COMPOUND  QUANTITIES 

31.  Addition  of  compound  numbers  is  the  process  of 
finding  the  sum  of  two  or  more  denominate  numbers  when 
one  or  more  of  them  is  compound. 

When  compound  numbers  are  to  be  added,  they  must  be  so 
placed  that  numbers  of  the  same  denomination  shall  stand  in  the 
same  vertical  column. 

Example  1. — Find  the  sum  of  4  lb.  13  oz.  14  dr.,  2  cwt.  93  lb.  15  oz. 
9  dr.,  and  3  T.  19  cwt.  35  lb.  12  oz.  11  dr.  (avoirdupois). 

Solution. — Arranging  the  numbers  as  explained  above, 


T. 

cwt. 

lb. 

OZ. 

dr. 

4 

13 

14 

2 

93 

15 

9 

3 

19 

35 

12 

11 

sum  4 

2 

34 

10 

2  Ans. 

Explanation. 

—  The 

column 

containing 

the  smallest 

denomination  is  first  added  and  equals  34  drams,  or  2  ounces 
and  2  drams.  Placing  the  2  drams  under  the  dram  column, 
add  the  2  ounces  in  with  the  ounce  column,  the  sum  of 
which  equals  42  ounces,  or  2  pounds  and  10  ounces.  Placing 
the  10  ounces  under  the  ounce  column,  add  the  2  pounds  in 
with  the  pound  column,  which  equals  134  pounds,  or  1  hun¬ 
dredweight  and  34  pounds.  Placing  the  34  pounds  under 
the  pound  column,  add  the  1  hundredweight  in  with  the  hun¬ 
dredweight  column,  which  equals  22  hundredweight,  or  1  ton 
and  2  hundredweight.  Placing  the  2  hundredweight  under 
the  hundredweight  column  and  adding  the  1  ton  in  with  the 
ton  column,  which  equals  4  tons,  the  final  sum  is  4  tons 
2  hundredweight  34  pounds  10  ounces  2  drams. 

Example  2. — Add  78  lb.  6  dr.,  85  lb.  9  oz.,  and  39  lb.  15  oz.  11  dr. 
(avoirdupois) . 

Solution. — Arranging  like  denominations  under  each  other, 

cwt.  lb.  oz.  dr. 

78  6 

85  9 

39  15  11 


sum  2 


3 


9 


1  Ans. 


§5 


ARITHMETIC 


13 


32.  To  add  compound  numbers: 

Rule. —  Write  the  given  numbers  so  that  numbers  denoting 
the  same  denomination  will  sta?id  in  the  same  column. 

Add  as  in  simple  addition  and  carry  from  the  sum  of  each 
denomination  one  for  as  many  units  as  it  takes  of  that  denomi¬ 
nation  to  make  a  unit  of  the  next  higher  denomination. 


EXAMPLES  FOR  PRACTICE 
Add  the  following  compound  numbers: 

(a)  3  lb.  24  oz.  3  dr.  and  7  lb.  11  oz.  14  dr.  (avoirdupois). 

( b )  3  gal.  2  qt.  1  pt.  3  gi.  and  21  gal.  3  qt.  1  pt.  3  gi.  (liquid 
measure) . 

( c )  56  T.  4  cwt.  75  lb.  14  oz.  and  48  T.  18  cwt.  37  lb.  8  dr. 
(avoirdupois) . 

( d )  12  oz.  13  pwt.  19  gr.  and  11  oz.  18  pwt.  21  gr.  (troy). 

( a )  12  lb.  4  oz.  1  dr. 

A  (6)  25  gal.  2  qt.  1  pt.  2  gi. 

(c)  105  T.  3  cwt.  12  lb.  14  oz.  8  dr. 

( d )  2  lb.  12  pwt.  16  gr. 


SUBTRACTION  OF  COMPOUND  QUANTITIES 

33.  Subtraction  of  compound  numbers  is  the  process  of 
finding  the  remainder,  or  difference,  between  two  denominate 
numbers  when  one  or  both  of  them  are  compound. 

Example  1. — From  2  lb.  7  oz.  5  dr.  2  sc.  12  gr.  subtract  1  lb.  9  oz. 
7  dr.  2  sc.  16  gr.  (apothecaries’  weight). 

Solution. —  lb.  oz.  dr.  sc.  gr. 

minuend  2  7  5  2  12 

subtrahend  1  9  7  2  16 

remainder  0  9  5  2  16  Ans. 

Explanation. — It  is  impossible  to  subtract  16  grains  from 
12  grains,  so  1  scruple,  or  20  grains,  is  borrowed  from  the 
scruple  column  making  32  grains,  from  which  16  grains  is 
subtracted,  leaving  16  grains  as  a  remainder.  In  the  same 
manner,  as  2  scruples  cannot  be  subtracted  from  1  scruple, 
1  dram,  or  3  scruples,  is  borrowed  from  the  dram  column 
making  4  scruples,  from  which  2  scruples  is  subtracted, 
leaving  2  scruples.  Similarly,  as  7  drams  cannot  be  sub¬ 
tracted  from  4  drams,  1  ounce,  or  8  drams,  is  borrowed  from 
the  ounce  column  making  12  drams,  from  which  7  drams  is 


14 


ARITHMETIC 


§5 


subtracted,  leaving  5  drams  as  a  remainder.  The  same 
method  is  followed  out  in  subtracting  the  ounces  and  the 
final  remainder  is  found  to  be  9  ounces  5  drams  2  scruples 
16  grains. 

Example  2. — Subtract  3  bu.  3  pk.  (5  qt.  from  7  bu.  3  pk.  3  qt.  (dry- 
measure)  . 


Solution. —  bu 

minuend  7 

subtrahend  3 

remainder  3 


pk. 

3 

3 

3 


qt. 

3 

J) 

5  Ans. 


34.  To  subtract  compound  numbers: 

Rule. —  Write  the  lesser  compound  number  under  the  greater 
so  that  numbers  of  the  same  denomination  zvill  be  in  the  same 
column  and  proceed  as  in  the  subtraction  of  simple  numbers. 

If  any  number  of  the  subtrahend  is  larger  than  the  correspond¬ 
ing  minuend  number ,  add  to  the  minuend  number  as  many  units 
as  make  one  of  the  next  higher  denomination  and  take  one  from 
the  minuend  number  of  such  higher  denomination. 


EXAMPLES  FOR  PRACTICE 


Subtract: 


(«) 

ib) 

(c) 

(d) 


15  oz.  from  1  lb.  3  oz.  (avoirdupois). 

7  lb.  5  oz.  3  dr.  from  14  lb.  3  oz.  2  dr.  (apothecaries’). 
1  lb.  10  oz.  16  pwt.  from  3  lb.  8  oz.  4  pwt.  (troy). 

44  lb.  3  oz.  10  dr.  from  11  cwt.  34  lb.  2  oz.  6  dr. 


Ans. 


(a)  4  oz. 

(b)  6  lb.  9  oz.  7  dr. 

(c)  1  lb.  9  oz.  8  pwt. 

(d)  10  cwt.  89  lb.  14  oz.  12  dr. 


MULTIPLICATION  OF  COMPOUND  QUANTITIES 

35.  Multiplication  of  compound  quantities  is  the  process 
of  adding  a  compound  number  to  itself  a  definite  number  of 
times. 

Example  1. — Multiply  3  gal.  2  qt.  1  pt.  3  gi.  (liquid  measure)  by  6. 


Solution. — 

gal. 

qt. 

pt. 

gi- 

multiplicand 

3 

2 

1 

3 

multiplier 

6 

product 

22 

1 

0 

2  Ans. 

§5 


ARITHMETIC 


15 


Explanation. — In  this  example  multiplying  3  gills  by  6, 
the  product  obtained  is  18  gills,  which  is  equal  to  4  pints  and 
2  gills.  Write  the  2  gills  in  the  gill  place  of  the  final 
product  and  add  the  4  pints  to  the  product  of  1  pint  multi¬ 
plied  by  6,  thus  1x6  =  6  and  6  -f  4  =  10.  The  product  of 
the  pints  is  equal  to  exactly  5  quarts,  so  a  cipher  is  written 
in  the  pint  place  of  the  final  product  and  the  5  quarts  is 
added  to  the  product  of  2  quarts  multiplied  by  6,  thus  2x6 
=  12  and  12  +  5  =  17.  The  product  of  the  quarts  is  equal 
to  4  gallons  and  1  quart.  Write  the  1  quart  in  the  quart 
place  of  the  final  product  and  add  the  4  gallons  to  the  product 
of  3  gallons  multiplied  by  6,  thus  3x6  =  18  and  18  +  4 
=  22.  As  there  is  no  higher  denomination  than  gallons  in 
the  example,  it  is  complete  and  the  final  product  is  22  gallons 
1  quart  2  gills. 

Example  2. — Multiply  35  bu.  3  pk.  6  qt.  (dry  measure)  by  9. 

Solution. —  bu.  pk.  qt. 

multiplicand  35  3  6 

multiplier  9 

product  323  1  6  A  ns. 

36.  To  multiply  compound  numbers: 

liule. — Multiply  each  denomination  of  the  multiplicand  by 
the  multiplier ,  as  in  the  multiplication  of  simple  numbers ,  and 
carry  as  in  the  addition  of  compound  numbers. 


EXAMPLES  FOR  PRACTICE 

Find  the  product  of  the  following  compound  numbers: 

( a )  5  gal.  2  qt.  1  pt.  (liquid  measure)  multiplied  by  3. 

(b)  13  cwt.  3  lb.  8  oz.  (avoirdupois)  multiplied  by  7. 

(c)  3  lb.  7  oz.  16  pwt.  (troy)  multiplied  by  4 

id)  25  T.  18  cwt.  66  lb.  12  oz.  (avoirdupois)  multiplied  by  6. 

' {a)  16  gal.  3  qt.  1  pt. 

(b)  91  cwt.  24  lb.  8  oz. 

(c)  14  lb.  7  oz.  4  pwt. 
.(d)  155  T.  12  cwt.  8  oz. 


Ans. 


16 


ARITHMETIC 


§5 


DIVISION  OF  COMPOUND  QUANTITIES 

37.  Division  of  compound  numbers  is  the  process  of 
dividing  a  compound  number  into  a  definite  number  of  equal 
parts. 

Example  1. — Divide  143  sq.  yd.  4  sq.  ft.  81  sq.  in.  (square  measure) 
by  5. 

Solution. —  sq.  yd.  sq.  ft.  sq.  in. 

divisor  5  )  143  4  81  dividend 

quotient  28  6  45  Ans. 

Explanation.— In  this  example,  143  square  yards  divided 
by  5  equals  28  square  yards  and  3  square  yards  left  over. 
Write  the  28  square  yards  in  the  quotient  and  reduce  the 
3  square  yards  to  square  feet,  adding  them  to  the  4  square 
feet  in  the  dividend;  thus,  3  square  yards  =  27  square  feet, 
adding  4  square  feet  =  31  square  feet,  which  when  divided 
by  5  equals  6  square  feet  and  1  square  foot  as  a  remainder. 
Reduce  the  1  square  foot  to  square  inches  and  add  them  to 
the  81  square  inches  in  the  dividend;  thus,  1  square  foot 
=  144  square  inches,  adding  81  square  inches  =  225  square 
inches,  which  when  divided  by  5  equals  45  square  inches. 
The  final  quotient  obtained  is  28  square  yards  6  square  feet 
45  square  inches. 

Example  2.  — Divide  323  bu.  1  pk.  6  qt.  (dry  measure)  by  9. 

Solution. —  bu.  pk.  qt. 

divisor  9  )  323  1  6  dividend 

quotient  35  3  6  Ans. 

38.  To  divide  compound  numbers: 

Rule. — Proceed  as  in  the  division  of  simple  numbers ,  dividing 
each  denomination  in  order  beginning  with  the  highest.  If  there 
is  a  remainder  after  dividing  any  denomination ,  reduce  it  to  the 
next  lower  denomination,  adding  in  the  number  of  this  denom¬ 
ination  in  the  dividend  if  any ,  and  proceed  as  before. 


§5 


ARITHMETIC 


17 


EXAMPLES  FOR  PRACTICE 


Divide: 

(a)  3  T.  13  cwt.  55  lb.  10  oz.  (avoirdupois)  by  6. 

(b)  72  sq.  yd.  4  sq.  ft.  31  sq.  in.  (square  measure)  by  7. 

(c)  36  cu.  yd.  1  cu.  ft.  960  cu.  in.  (cubic  measure)  by  8. 

( d )  21  T.  5  cwt.  52  lb.  (a  oirdupois)  by  8. 

(a)  12  cwt.  25  lb.  15  oz. 


Ans. 


(b)  10  sq.  yd.  3  sq.  ft.  25  sq.  in. 
\c)  4  cu.  yd.  13  cu.  ft.  1,200  cu. 
(d)  2  T.  13  cwt.  19  lb. 


in. 


MENSURATION 


MENSURATION  OF  SURFACES 


DEFINITIONS 

1.  Mensuration  treats  of  the  measurement  of  lines , 
angles,  surfaces,  and  solids. 

2.  A  line  expresses  length  or  distance  without  breadth 
or  thickness. 


3.  A  straight  line,  Fig.  1,  is  one  that 
does  not  change  its  direction  throughout  its  fxG.  i 
whole  length. 


4.  A  curved  line,  Fig.  2,  changes  its 
direction  at  every  point. 


Fig.  2 


5.  Parallel  lines,  Fig.  3,  are  those 
that  are  equally  distant  from  each  other  at 
all  points. 


6.  A  line  is  perpendicular  to  another, 

Fig.  4,  when  it  meets  that  line  so  as  not  to 

incline  toward  it  on  either  side.  _ 

Fig.  4 


7.  A  vertical  line,  Fig.  5,  is  one  that 
points  toward  the  center  of  the  earth;  it  is 
also  known  as  a  plumb-line. 

8.  A  horizontal  line,  Fig.  5,  is  one 
that  makes  a  right  angle  with  a  vertical  line. 


8 

f 

g 


Horizontal 

Fig.  5 


9.  A  surface  is  that  which  has  length  and  breadth 
without  thickness. 


For  notice  of  copyright,  see  page  immediately  following  the  title  page 

$6 


2 


MENSURATION 


§6 


10.  A  plane  surface  is  one  in  which  if  two  points  be 
taken,  a  straight  line  connecting  them  will  be  wholly  in  the 
surface. 

11.  A  curved  surface  is  one  no  part  of  which  is 
plane. 

12.  An  angle  is  the  inclination  of  two  lines  one  to  the 
other  and  is  measured  in  degrees  (°). 

13.  A  riglit  angle,  Figs.  4  and  5,  is  formed  by  two 
lines  that  are  perpendicular  to  each  other,  and  is  90°  in 
magnitude. 

14.  An  acute  angle,  Fig.  6,  is  an  angle  of  less  than  90°. 


Fig.  6  Fig.  7 

15.  An  obtuse  angle,  Fig.  7,  is  an  angle  of  more 
than  90°. 

16.  A  plane  figure  is  any  part  of  a  plane  surface 
bounded  by  straight  or  curved  lines. 

17.  The  ai’ea  of  a  plane  figure  is  its  surface  contents. 


TRIANGLES 


18.  A  triangle  is  a  plane  figure  bounded  by  thrde 
straight  lines  and  having  three  angles. 


19.  The  altitude  of  a  triangle  is  the 
distance  from  its  apex  to  base  measured 
perpendicularly  to  the  base.  In  the  tri¬ 
angle  a  be,  Fig.  8,  the  dotted  line  b  d repre¬ 
sents  the  altitude  of  the  triangle,  while 
the  base  of  the  triangle  is  represented  by 
the  line  ac. 


20.  An  equilateral  triangle,  Fig.  9,  is  one  that  has 
all  of  its  sides  equal  and  each  of  its  angles  of  60°  magnitude. 


§6 


MENSURATION 


3 


21.  An  isosceles  triangle, 
equal  sides  and  two  equal 
angles. 

22.  A  scalene  triangle, 
Fig.  11,  is  one  that  has  all  of 
its  sides  and  all  of  its  angles 
unequal. 


Fig.  10,  is  one  that  has  two 


23.  A  right  tri¬ 
angle,  Fig.  12,  is  one  that 
has  one  angle  a  right 
angle. 

24.  To  find  the  area 
of  a  triangle: 


Fig.  11 


Rule. — Multiply  the  base  by  the  altitude  and  divide  the 
product  by  2. 

Example. — The  base  of  a  triangle  is  14  inches  in  length  and  the 
altitude  is  12  inches;  what  is  the  area? 

14  in.  X  12  in.  . 

Solution. —  - 2 -  =  °4  sq.  in.  Ans. 

Note.— In  the  above  example  it  will  be  noticed  that  by  multiplying:  inches  by 
inches  the  product  obtained  is  square  inches;  similarly,  feet  multiplied  by  feet  or 
rods  by  rods  equals  square  feet  or  square  rods,  etc.  It  must  be  remembered  that 
only  like  numbers  can  be  multiplied  together  and  that  feet  can  never  be  multiplied 
by  inches,  nor  rods  by  feet;  consequently,  in  all  problems  dealing  with  mensuration, 
all  dimensions  must  be  reduced  to  like  terms  before  multiplying. 


EXAMPLES  FOR  PRACTICE 


Find  the  areas  of  the  following  triangles,  the  length  of  the  base  and 
altitude  being  respectively: 

(a)  10  inches  and  8  inches.  (a) 

(b)  34  feet  and  42  feet. 

( c )  114  inches  and  212  inches. 

(d)  34  miles  and  18  miles. 


Ans. 


W 

(c) 

(d) 


40  sq.  in. 

714  sq.  ft. 
12,084  sq.  in. 
306  sq.  mi. 


25.  To  find  tlie  area  of  a  triangle  when  the  alti¬ 
tude  is  unknown  but  the  length  of  each  side  is  given: 

Rule. — From  one-half  the  sum  of  the  three  sides ,  subtract 
each  of  the  sides  separately  and  multiply  the  remainders  together 
and  by  one-half  the  sum  of  the  sides;  the  square  root  of  the 
Product  will  be  the  area  of  the  triangle. 


4 


MENSURATION 


6 


Example. — What  is  the  area  of  a  triangle  the  sides  of  which  are, 
respectively,  16,  16,  and  12  feet  in  length? 

Solution.—  16  +  16  +  12  =  44;  44  -5-  2  -  22;  22  -  16  =  6; 

22  —  16  =  6;  22  -  12  =  10; 

6  X  6  X  10  X  22  =  7,920;  V L920  =  88.99  sq.  ft.  Ans. 


EXAMPJ.ES  for  practice 

1.  Find  the  area  of  a  triangle  the  sides  of  which  are,  respectively, 

32,  32,  and  24  inches  in  length.  Ans.  355.977  sq.  in. 

2.  Find  the  area  of  a  triangle  the  sides  of  which  are,  respectively, 

18,  24,  and  22  feet  in  length.  Ans.  189.314  sq.  ft. 


QUADRILATERALS 

26.  A  quadrilateral  is  a  plane  figure  bounded  by  four 
straight  lines. 


27.  A  parallelogram  is 

sides  of  which  are  parallel. 


a  quadrilateral  the  opposite 

28.  A  rectangle, 
Fig.  18,  is  a  parallelogram 
having  all  of  its  angles 
right  angles. 


29.  A  square.  Fig. 
14,  is  a  parallelogram  hav¬ 


ing  all  of  its  angles  right  angles  and  all  of  its  sides  of  equal 
length. 


30.  A  rhomboid,  Fig.  15,  is  a  parallelogram  having 
none  of  its  angles  right  angles. 


Fig.  15 


31.  A  rhombus,  Fig.  16,  is  a  parallelogram  having  all 
of  its  sides  of  equal  length  but  none  of  its  angles  right 
angles. 


§6 


MENSURATION 


5 


32.  The  altitude  of  a  parallelogram  is  the  distance 
between  two  opposite  sides  measured  perpendicularly,  as 
indicated  by  the  dotted  lines  in  Figs.  15  and  16. 

33.  To  fiud  the  area  of  a  parallelogram: 

Rule. — Multiply  the  altitude  by  the  base  and  the  product  will 
be  the  area. 

Example. — Find  the  area  of  a  parallelogram  the  base  of  which  is 
345  inches  and  the  altitude  423  inches. 

Solution.—  423  in.  X  345  in.  =  145,935  sq.  in.  Ans. 


EXAMPLES  FOR  PRACTICE 


Find  the  areas  of  the  following  parallelograms,  the  lengths  of  the 
bases  and  altitudes  being  respectively: 


(a)  145  inches  and  136  inches. 

(b)  2,034  feet  and  23  feet. 

(c)  135  rods  and  4|  rods. 

(d)  39  feet  and  14^  feet. 


Ans.< 


' (a)  19,720  sq.  in. 

(b)  46,782  sq.  ft. 

( c )  567  sq.  rd. 

.  ( d )  559  sq.  ft. 


34.  A  trapezoid,  Fig.  17,  is  a 

quadrilateral  having  only  two  of 
its  sides  parallel.  fig.  17 

35.  The  altitude  of  a  trapezoid  is  always  measured 
perpendicularly  between  the  parallel  sides,  as  shown  by  the 
dotted  line  in  Fig.  17. 

36.  To  find  the  area  of  a  trapezoid: 

Rule. — Multiply  one-half  the  sum  of  the  parallel  sides  by  the 
altitude. 

Example. — The  parallel  sides  of  a  trapezoid  are,  respectively,  12 
and  28  feet  in  length,  and  the  altitude  is  30  feet;  what  is  the  area  of 
the  figure? 

Solution. —  12  ft.  +  28  ft.  =  40  ft.;  40  ft.  -p  2  =  20  ft. 

20  ft.  X  30  ft.  =  600  sq.  ft.  Ans. 


EXAMPLES  FOR  PRACTICE 

1.  What  is  the  area  of  a  trapezoid  the  parallel  sides  of  which  are, 
respectively,  54  and  78  feet  in  length,  and  the  altitude  of  which  is 
64  feet?  Ans.  4,224  sq.  ft. 


6 


MENSURATION 


§6 


2.  What  is  the  area  of  a  trapezoid  the  parallel  sides  of  which  are, 
respectively,  8  and  18  inches  in  length,  and  the  altitude  of  which  is 
23  inches?  Ans.  299  sq.  in. 


37.  A  trapezium,  Fig.  18,  is  a  quadrilateral  that  has 
no  two  sides  parallel. 

38.  A  line  joining  two  oppo¬ 
site  corners  of  a  quadrilateral,  as 
for  instance  the  line  a  b,  Fig.  18, 
is  known  as  a  diagonal. 

39.  To  find  the  area  of  a 
trapezium: 

b  Rule. — Divide  the  figure  into 

tzvo  triangles  by  ?neans  of  a  diagonal; 
the  sum  of  the  areas  of  these  triangles  equals  the  area  of  the 
trapezium. 

Example. — What  is  the  area  of  a  trapezium  whose  diagonal  is 
43  inches  long,  the  length  of  the  perpendicular  lines  dropped  on  the 
diagonal  from  the  opposite  corners  being  22  and  26  inches,  respectively? 

Note.— The  perpendicular  lines  drawn  from  opposite  corners  of  a  quadrilateral 
to  its  diagonal  constitute  the  altitudes  of  the  two  triangles  into  which  the  diagonal 
divides  the  quadrilateral.  Thus,  in  Fig.  18,  the  line  f  d  represents  the  altitude  of  the 
triangle  a  db,  and  the  line  ec  the  altitude  of  the  triangle  ac  b. 

Solution. —  43  in.  X  22  in.  =  946  sq.  in.;  946  sq.  in.  -p  2  =  473  sq. 
in.,  area  of  one  triangle;  43  in.  X  26  in.  =  1,118  sq.  in.;  1,118  sq.  in. 
ri-  2  =  659  sq.  in.,  area  of  other  triangle. 

473  sq.  in.  +  559  sq.  in.  =  1,032  sq.  in.,  area  of  trapezium.  Ans. 


a 


EXAMPLES  FOR  PRACTICE 

1.  The  diagonal  of  a  trapezium  is  26  feet  in  length  and  the  per¬ 

pendiculars  from  the  opposite  corners  to  the  diagonal  are  8  and  14  feet, 
respectively,  in  length;  what  is  the  area?  Ans.  286  sq.  ft. 

2.  The  diagonal  of  a  trapezium  is  78  inches  and  the  perpendicu¬ 

lars  42  and  48  inches,  respectively,  in  length;  what  is  the  area  of  the 
figure?  Ans.  3,510  sq.  in. 


§6 


MENSURATION 


7 


POLYGONS 

40.  A  polygon  is  a  plane  figure  bounded  by  straight 
lines.  The  term  is  usually  applied  to  a  figure  having  more 
than  four  sides.  The  bounding  lines  are  called  the  sides, 
and  the  sum  of  the  lengths  of  all  the  sides  is  called  the  per¬ 
imeter  of  the  polygon. 

41.  A  regular  polygon  is  one  in  which  all  the  sides 
and  all  the  angles  are  equal. 

42.  A  polygon  of  five  sides  is  called  a  pentagon;  one 
of  six  sides,  a  hexagon;  one  of  seven  sides,  a  heptagon, 
etc.  Regular  polygons  having  from  five  to  twelve  sides 
are  shown  in  Fig.  19. 


Pentagon  Hexagon  Heptagon  Octagon  Decagon  Dodecagon 

Fig.  19 


43.  To  find  the  area  of  a  regular  polygon: 

Rule. — Multiply  the  perimeter  by  one-half  the  length  of  the 
perpendicular  from  its  center  to  one  of  its  sides. 

Example. — The  perimeter  of  a  regular  polygon  is  28  inches  in  length 
and  the  perpendicular  distance  from  its  center  to  one  side  is  8  inches; 
what  is  its  area? 

Solution. —  8  in.  -f-  2  =  4  in.;  28  in.  X  4  in.  =  112  sq.  in.  Ans. 


EXAMPLES  FOR  PRACTICE 

1.  If  the  perimeter  of  a  regular  polygon  is  78  feet  in  length  and  the 

distance  from  its  center  to  one  side  measured  perpendicularly  is  21  feet, 
what  is  its  area?  Ans.  819  sq.  ft. 

2.  The  perimeter  of  a  regular  polygon  is  112  inches  in  length  and 

the  perpendicular  distance  from  the  center  to  one  side  is  32  inches; 
what  is  the  area?  Ans.  1,792  sq.  in. 


8 


MENSURATION 


§6 


THE  CIRCLE 

44.  A  circle,  Fig.  20,  is  a  plane  figure  bounded  by  a 
curved  line,  called  the  circumference,  every  portion  of 
which  is  equally  distant  from  a  point  within  called  the  center. 


45.  The  diameter  of  a  circle  is  any  straight  line  drawn 
through  its  center  and  terminating  at  each  end  in  the  cir¬ 
cumference.  Thus  the  line  a  b,  Fig.  21,  is  a  diameter  of 
the  circle. 

46.  If  a  circle  is  divided  into  halves,  each  half  is  called 
a  semi-circle,  and  each  half  of  the  circumference  is  called  a 
semi-circumference. 


47.  Any  straight  line  terminating  at  each  end  in  the 
circumference  but  not  passing  through  the  center  is  called 


a  chord,  as  for  instance 
the  line  a  e.  Fig.  22. 


Fig.  23  Fig.  24 

49.  An  arc  of  a  circle  (see 
its  circumference. 


48.  A  straight  line 
drawn  from  the  center  to 
the  circumference  of  a 
circle  (as  a  c,  Fig.  23)  is 
called  a  radius. 

a  d  e,  Fig.  24)  is  any  part  of 


50.  To  find  the  circumference  of  a  circle: 

Rule. — Multiply  the  diameter  by  3.1416. 

Note. —  3.1416  is  the  approximate  length  of  the  circumference  of 
a  circle  whose  diameter  is  1 . 

Example.— What  is  the  circumference  of  a  circle  the  diameter  of 
which  is  48  inches? 

Solution. —  48  in.  X  3.1416  =  150.7968  in.  Ans. 


6 


MENSURATION 


9 


51.  To  find  the  diameter  of  a  circle  with  a  given 
length  of  circumference: 

Rule. — Divide  the  circumference  by  3.1416. 

Example. — What  is  the  diameter  of  a  circle  the  length  of  circum¬ 
ference  of  which  is  8  feet? 

Solution. —  8  ft.  -r-  3.1416  =  2.5465  ft.  Ans. 


EXAMPLES  FOR  PRACTICE 


Find  the  circumferences  of  circles  having  the  following  diameters: 


(а)  24  inches. 

(б)  35  feet. 

( c )  27  rods. 

(a)  79  yards. 


Ans. 


f (a)  75.3984  in. 
(b)  109.956  ft. 
(0  84.8232  rd. 
l(rf)  248.1864  yd. 


52.  To  find  the  area  of  a  circle: 

Rule. — Multiply  the  square  of  the  diameter  by  .7854. 

Note. —  .7854  is  the  area  of  a  circle  whose  diameter  is  1. 

Example. — What  is  the  area  of  a  circle  the  diameter  of  which  is 
75  inches? 

Solution. —  75  in.  X  75  in.  X  .7854  =  4,417.875  sq.  in.  Ans. 


EXAMPLES  FOR  PRACTICE 


Find  the  areas  of  circles  of  the  following  diameters: 


(a)  22  inches. 

(b)  47  yards. 

( c )  768  rods. 

(d)  176  inches. 


Ans. 


(a)  380.1336  sq.  in. 

{d)  1,734.9486  sq.  yd. 

'  (c)  463,247.7696  sq.  rd. 
.(d)  24,328.5504  sq.  in. 


53.  To  find  the  length  of  one  side  of  a  square 
inscribed  in  a  given  circle: 

Rule. — Multiply  the  diameter  of  the  circle  by  .707107 '. 

Note. — A  square  is  said  to  be  inscribed  in  a  circle  when  the  vertices 
of  all  its  angles  lie  in  the  circumference  of  the  circle.  .707107  is  the 
length  of  the  side  of  a  square  inscribed  in  a  circle  whose  diameter  is  1. 

Example. — How  thick  is  the  largest  square  stick  that  can  be  inserted 
in  a  pipe  5  inches  in  diameter  on  the  inside? 

Solution. —  5  in.  X  .707107  =  3.5355  in.  Ans. 


10 


MENSURATION 


§6 


EXAMPLES  FOR  PRACTICE 

1 .  What  is  the  thickness  of  the  largest  square  iron  rod  that  will  just 

fit  in  a  round  hole  li  inches  in  diameter?  Ans.  1.06  in. 

2.  What  is  the  thickness  of  a  square  stick  of  timber  that  may  be 

cut  from  a  log  28  inches  in  diameter?  Ans.  19.7989  in. 


54.  To  find  the  length  of  one  side  of  a  square 
equal  in  area  to  a  given  circle: 

Rule. — Multiply  the  diameter  of  the  circle  by  .886227. 

Note. —  .886227  is  the  length  of  the  side  of  a  square  equal  in  area 

to  a  circle  whose  diameter  is  1. 

Example. — What  is  the  length  of  one  side  of  a  square  that  is  equal 
in  area  to  a  circle  15  inches  in  diameter? 

Solution. —  15  in.  X  .886227  =  13.293  in.  Ans. 


EXAMPLES  FOR  PRACTICE 

1.  What  is  the  length  of  one  side  of  a  square  that  will  have  an  area 

equal  to  that  of  a  piston  20  inches  in  diameter?  Ans.  17.7245  in. 

2.  What  is  the  length  of  one  side  of  a  square  field  that  will  contain 
the  same  number  of  acres  as  a  circular  field  700  feet  in  diameter? 

Ans.  620.3589  ft. 


MENSURATION  OF  SOLIDS 


THE  PRISM 

55.  A  solid,  or  solid  body,  is  one  that  has  three 
dimensions;  viz.,  length,  breadth,  and  thickness. 

56.  A  prism  is  a  solid  body  the  ends  of  which  are 
formed  by  two  similar  plane  figures  that  are 
equal  and  parallel  to  each  other,  and  whose 
sides  are  parallelograms.  Prisms  are  triangular, 
rectangular,  square,  etc.  according  to  the  char¬ 
acter  of  the  figure  forming  the  ends. 

57.  A  parallelopipedon,  Fig.  25,  is  a 
prism  whose  bases  (ends)  are  parallelograms. 


A 

/ 

! 

i 

j 

i 

i 

i 

i 

■ 

7 

Fig.  25 


6 


MENSURATION 


11 


58.  A  cube,  Fig.  26,  is  a  prism  whose  faces  and  ends 
are  squares.  All  the  faces  of  a  cube  are  equal. 

59.  In  the  case  of  plane  figures,  perimeters 
and  areas  must  be  considered.  In  the  case  of 
solids,  the  areas  of  their  outside  surfaces  and 
their  contents  or  volumes  must  be  considered. 


Fig.  2f> 


60.  The  base  of  a  prism  is  either  end,  and  of  solids  in 
general,  the  ends  on  which  they  are  supposed  to  rest. 

61.  To  find  the  surface  area  of  a  prism: 


Rule. — Multiply  the  length  of  the  perimeter  of  the  base  by  the 
altitude,  and  to  the  product  add  the  area  of  both  ends. 

Example. — What  is  the  surface  area  of  a  square  prism  the  base  of 
which  is  14  inches  square  and  the  altitude  25  inches  in  length? 

Solution. —  14  in.  X  4  =  56  in.,  perimeter  of  base 

56  in.  X  25  in.  =  1,400  sq.  in.,  area  of  sides 
14  in.  X  14  in.  =  196  sq.  in.,  area  of  one  base 
196  sq.  in.  X  2  =  392  sq.  in.,  area  of  both  bases 
1,400  sq.  in.  +  392  sq.  in.  =  1,792  sq.  in.,  total  surface  area.  Ans. 


EXAMPLES  FOR  PRACTICE 

1.  Find  the  surface  area  of  a  rectangular  prism  whose  base  is 
10  inches  long  and  12  inches  wide  and  which  is  13  inches  high. 

Ans.  812  sq.  in. 

2.  What  is  the  surface  area  of  a  square  prism  the  base  of  which 
is  175  inches  square  and  whose  altitude  is  342  inches? 

Ans.  300,650  sq.  in. 


62.  To  find  tlie  contents  or  volume  of  a  prism 
or  rectangular  box: 

Rule. — Multiply  the  width  by  the  depth  and  by  the  length; 
or  find  the  area  of  the  base  according  to  the  rule  previously 
given ,  which  when  multiplied  by  the  height  equals  the  contents 
or  solidity  of  the  prism. 

Example.— What  is  the  capacity  of  a  box  36  inches  long,  the 
ends  being  14  inches  by  28  inches? 

Solution. —  28  in.  X  Min.  X  36  in.  =  14,112  cu.  in.  Ans. 


12 


MENSURATION 


§6 


Note.— It  lias  been  stated  that  inches  multiplied  by  inches  equals  square  inches 
or,  similarly,  yards  multiplied  by  yards  equals  square  yards.  Continuing:  still  further, 
as  is  necessary  in  finding:  the  contents,  volume,  solidity,  or  capacity  of  solids:  square 
inches  or  square  yards  multiplied  by  inches  or  yards  equals  cubic  inches  or  cubic 
yards,  etc. 

From  this  it  will  be  seen  that  by  multiplying:  together  the  two  dimensions  of  a 
surface,  such  as  a  rectangle,  the  area  of  the  figure  will  be  expressed  in  square  units, 
while  if  the  three  dimensions  of  a  solid,  as  for  instance,  a  parallelopipedon.  are  multi¬ 
plied  together  the  contents,  or  solidity,  of  the  solid  is  expressed  in  cubical  units. 


EXAMPLES  FOR  PRACTICE 

1.  What  is  the  capacity  of  a  box  the  ends  of  which  are  24  inches 

square  and  the  length  of  which  is  44  inches?  Ans.  25,344  cu.  in. 

2.  What  is  the  capacity,  in  cubic  feet,  of  a  freight  car  34  feet  long, 

8  feet  wide,  and  7  feet  high?  Ans.  1,904  cu.  ft. 

3.  How  many  cubic  inches  are  there  in  a  stick  of  timber  8^  feet 

long,  7  inches  wide,  and  4^  inches  thick?  Ans.  3,213  cu.  in. 

Suggestion.— Reduce  the  8i  feet  to  inches. 

4.  What  is  the  contents  of  a  tank  13  feet  square  and  12  feet  high? 

Ans.  2,028  cu.  ft. 

5.  How  many  cubic  feet  are  there  in  a  cube  whose  sides  are  12  feet 

in  length?  Ans.  1,728  cu.  ft. 


THE  CYLINDER 

A  cylinder,  Fig.  27,  is  a  body  of  uniform  diameter 
the  ends,  or  bases,  of  which  are  equal  parallel 
circles. 

64.  To  find  the  surface  area  of  a  cylinder: 

Rule.—  Multiply  the  circumference  of  the  base  by 
the  height  of  the  cylinder  and  to  this  product  add  the 
area  of  the  ends. 

Example. — What  is  the  surface  area  of  a  cylinder  6  inches  in 
diameter  and  13  inches  high? 

Solution. — 

62  X  .7854  =  28.2744  sq.  in.,  area  of  one  end 
28.2744  sq.  in.  X  2  =  56.5488  sq.  in.,  area  of  both  ends 
6  in.  X  3.1416  =  18.8496  in.,  length  of  circumference 
18.8496  X  13  =  245.0448  sq.  in.,  area  of  convex  surface 
245.0448  +  56.5488  =  301.5936  sq.  in.,  total  surface  area.  Ans. 

Note.— The  convex  surface  of  a  solid  is  the  curved  surface;  thus,  the  area  of  the 
convex  surface  of  a  cylinder  is  its  total  surface  area  less  the  area  of  the  ends. 


63. 


Fig.  27 


6 


MENSURATION 


13 


EXAMPLES  FOR  PRACTICE 

1.  What  is  the  surface  area  of  a  cylinder  7  feet  long  and  21  inches 
in  diameter?  Ans.  6,234.5052  sq.  in. 

2.  What  is  the  surface  area  of  a  cylinder  21  feet  long  and  27  inches 

in  diameter?  Ans.  22,520.5596  sq.  in. 

3.  What  is  the  surface  area  of  a  21-foot  boiler,  6  feet  in  diameter? 

Neglect  the  curvature  of  the  heads.  Ans.  452.3904  sq.  ft. 


65.  To  find  the  contents  or  volume  of  a  cylinder: 

Rule. — First  find  the  area  of  the  base  according  to  the  rule  in 
Art.  52,  and  then  multiply  the  area  of  the  base  by  the  altitude. 

Example. — How  many  cubic  feet  of  water  will  a  cylindrical  tank 
12  feet  in  diameter  and  14  feet  high  hold? 

Solution. —  122  X  .7854  =  113.0976  sq.  ft.,  area  of  base;  113.0976 
sq.  ft.  X  14  ft.  =  1,583.3664  cu.  ft.  Ans. 


EXAMPLES  FOR  PRACTICE 

1.  What  is  the  contents  of  a  cylinder  35  inches  in  diameter  and 

4|  feet  high?  Ans.  51,954.21  cu.  in. 

2.  What  is  the  capacity  of  a  cylindrical  tank  8  feet  in  diameter  and 

9  feet  deep?  Ans.  452.3904  cu.  ft. 

3.  If  the  water  in  a  circular  tank  44  inches  in  diameter  is  27  inches 
deep,  how  many  cubic  feet  of  water  are  there  in  the  tank? 

Ans.  23.7583  cu.  ft. 

4.  How  many  cubic  inches  in  a  cylinder  19  inches  long  and  11  inches 

in  diameter?  Ans.  1,805.6346  cu.  in. 

THE  PYRAMID  AND  CONE 

66.  A  pyramid,  Fig.  28,  is  a  solid  the  base  of  which 
is  a  polygon  and  the  sides  of 
which  taper  uniformly  to  a 
point  called  the  apex,  or 
vertex. 

67.  A  cone,  Fig.  29,  is 
a  solid  having  a  circle  as  a 
base  and  a  convex  surface 
tapering  uniformly  to  a  point  called  the  apex,  or  vertex. 


14  MENSURATION  §6 

68.  The  altitude  of  a  pyramid  or  cone  is  the  perpen¬ 
dicular  distance  from  the  vertex  to  the  base. 

69.  To  find  tlie  contents  ox*  volume  of  a  cone  or 
pyramid: 

Rule  .—Multiply  the  area  of  the  base  by  one-third  the  altitude. 

Example. — What  is  the  solid  contents  of  a  cone  30  feet  high  and 
5  feet  in  diameter  at  the  base? 

Solution. —  5s  X  .7854  =  19.635  sq.  ft.  area  of  base; 

i  of  30  ft.  =  10  ft. 

19.635  sq.  ft.  X  10  ft.  =  196.35  cu.  ft.  Ans. 


EXAMPLES  FOR  PRACTICE 

1.  Find  the  contents  of  a  cone  39  inches  high  and  12  inches  in 

diameter  at  the  base.  Ans.  1,470.2688  cu.  in. 

2.  Find  the  contents  of  a  square  pyramid  300  feet  high  and  325  feet 

square  at  the  base.  Ans.  10,562,500  cu.  ft. 


70. 


THE  FRUSTUM  OF  A  PYRAMID  OR  CONE 

If  a  pyramid  be  cut  by  a  plane  parallel  to  the  base, 

so  as  to  form  two  parts,  as  in 
Fig.  30,  the  lower  part  is  called 
the  frustum  of  the  pyramid. 

If  a  cone  be  cut  in  a  similar  man¬ 
ner,  as  in  Fig.  31,  the  lower  part 
is  called  the  frustum  of  the  cone. 


71.  To  find  tlie  contents  or 
volume  of  the  frustum  of  a 
pyramid  oi*  cone: 


Rule. — Find  the  areas  of  the  two  ends  of  the  frustum;  multiply 
them  together  and  extract  the  square  root  of  the  product.  To 
the  result  thus  obtained  add  the  two  areas  and  multiply  the  sum 
by  one-third  of  the  altitude. 

Example. — What  is  the  capacity  of  a  tank  shaped  like  the  frustum 
of  a  cone,  the  inside  diameter  of  the  top  being  10  feet  and  of  the 
bottom  14  feet,  and  the  depth  of  the  tank  being  12  feet? 


§6 


MENSURATION 


15 


Solution. —  10  ft.  X  10  ft.  X  .7854  =  78.54  sq.  ft.,  area  of  small 
end;  14  ft.  X  14  ft.  X  .7854  =  153.9384  sq.  ft.,  area  of  large  end; 
153.9384  X  78.54  =  12,090.321936;  Vl2, 090.321936  =  109.956  sq.  ft.; 
109.956  +  153.9384  +  78.54  =  342.4344  sq.  ft.;  12  ft.  -p  3  =  4  ft. 
342.4344  sq.  ft.  X  4  ft.  =  1,369.7376  cu.  ft.  Ans. 


EXAMPLES  FOR  PRACTICE 

1.  What  is  the  volume  of  the  frustum  of  a  square  pyramid  the 

length  of  which  is  30  feet,  the  top  being  10  feet  square  and  the  bottom 
20  feet  square?  Ans.  7,000  cu.  ft. 

2.  What  is  the  contents  of  a  round  stick  of  timber  20  feet  long, 
1  foot  in  diameter  at  the  larger  end,  and  \  foot  at  the  small  end? 

Ans.  9.162  cu.  ft. 


THE  SPHERE 

72.  A  sphere,  Fig.  32,  is  a  solid  bounded  by  a  continuous 
convex  surface,  every  part  of  which  is 
equally  distant  from  a  point  within  called 
the  center. 

73.  The  diameter,  or  axis,  of  a  sphere 
is  a  line  passing  through  its  center  and  ter¬ 
minating  at  each  end  at  the  surface. 

74.  To  find  the  surface  area  of  a  sphere: 

Rule. — Square  the  diameter  and  multiply  the  result  by  3.1416. 

Example. — What  is  the  surface  area  of  a  sphere  14  inches  in 
diameter? 

Solution. — 

143  X  3.1416  =  14  X  14  X  3.1416  =  615.75  sq.  in.  Ans. 


EXAMPLES  FOR  PRACTICE 

1.  What  is  the  surface  area  of  a  sphere  10  inches  in  diameter? 

Ans.  314.16  sq.  in. 

2.  The  diameter  of  the  earth  is  about  8,000  miles;  what  is  its 

approximate  surface  area?  Ans.  201,062,400  sq.  mi. 

75.  To  find  the  contents  or  volume  of  a  sphere: 

Rule. — Multiply  the  ciibe  of  the  diameter  by  .5236. 


Fig.  32 


16  MENSURATION  §6 

Example. — How  many  cubic  inches  of  ivory  in  a  billiard  ball 
2  inches  in  diameter? 

Solution. —  23  X  .5236  =  4.1888  cu.  in.  Ans. 

EXAMPLES  FOR  PRACTICE 

1.  How  many  cubic  inches  are  there  in  a  sphere  20  inches  in 

diameter?  Ans.  4,188.8  cu.  in. 

2.  What  is  the  contents  of  a  spherical  float  9  inches  in  diameter? 

Ans.  381.7044  cu.  in. 


MENSURATION  OF  LUMBER 

76.  Lumber  is  measured  by  board  measure,  which  is 
an  adaptation  of  square  measure. 

77.  A  board  foot  is  considered  as  1  square  foot  of  board 
1  inch  thick:  therefore  1,000  feet  of  lumber  is  equal  to 
1,000  square  feet  of  boards  1  inch  thick. 

78.  To  find  the  number  of  feet  of  lumber  in  1-incli 
boards: 

Rule. — Multiply  the  length  of  the  board ,  in  feet ,  by  the  width , 
in  inches ,  and  divide  the  product  by  12. 

Example. — How  many  feet  of  lumber  are  there  in  a  1-inch  board 
18  feet  long  and  8  inches  wide? 

Solution. —  — —  =  12  ft.  Ans. 


EXAMPLES  FOR  PRACTICE 

1.  How  many  board  feet  in  a  1-inch  board  21  feet  long  and 

18  inches  wide?  Ans.  31 J  ft. 

2.  How  many  board  feet  in  a  1-inch  board  12  feet  long  and 

24  inches  wide?  Ans.  24  ft. 


79.  To  find  the  number  of  feet  of  lumber  in 
joists,  beams,  etc.: 

Rule. — Multiply  the  tvidth,  in  inches ,  by  the  thickness ,  in 
inches ,  and  by  the  length,  in  feet.  Divide  this  product  by  12 
and  the  quotient  is  the  number  of  feet  of  lumber  in  the  stick. 


§6 


MENSURATION 


17 


Example. — How  many  feet  of  lumber  in  a  joist  4  inches  wide, 
3  inches  thick,  and  12  feet  long? 

„  4X3X12  10£l  . 

Solution.—  - -  =  12  ft.  Ans. 


EXAMPLES  FOR  PRACTICE 

1.  How  many  feet  of  lumber  in  a  beam  8  inches  wide,  3  inches 

thick,  and  16  feet  long?  Ans.  32  ft. 

2.  How  many  feet  of  lumber  in  a  timber  12  inches  wide,  16  inches 

deep,  and  24  feet  in  length?  Ans.  384  ft. 


MACHINE  ELEMENTS 


SHAFTING 


CLASSES  OF  SHAFTING 

1.  A  shaft  may  be  defined  as  a  rod,  generally  of  iron  or 
steel,  that  may  be  rotated  for  the  purpose  of  transmitting 
power  in  a  certain  direction  and  quantity,  depending  largely 
on  the  connections  of  the  driven  pieces  with  the  shaft  and  its 
size.  The  word'  shafting,  as  used  in  the  mill,  generally 
includes  only  that  which  is  employed  in  the  main  transmis¬ 
sion  of  power  to  various  machines,  although  the  term  actually 
includes  parts  of  a  machine  that  are  of  similar  construction. 

The  size  of  a  shaft  is  determined  by  its  diameter,  which  is 
expressed  in  inches  and  fractional  parts  of  an  inch.  It  is 
always  better  to  use  a  shaft  of  sufficient  diameter  to  transmit 
the  required  horsepower  easily  than  to  place  the  strain  on  a 
very  small  shaft  and  run  the  risk  of  its  breaking. 

Formerly  wrought-iron  shafts  were  largely  used,  but  these 
are  being  replaced  by  turned  or  cold-rolled  steel  shafting. 
Large  shafts,  such  as  are  employed  for  flywheels  and  shafts 
where  a  considerable  amount  of  power  is  transmitted,  are 
generally  forged  from  the  best  quality  of  steel. 

Shafting  of  almost  any  diameter  and  of  nearly  any  desired 
length  may  be  purchased  to  suit  circumstances;  the  freight, 
however,  is  higher  on  pieces  over  a  certain  length. 

For  notice  of  copyright ,  see  page  immediately  following  the  title  page 

27 


2 


MECHANICAL  DEFINITIONS 


§7 


2.  The  shafting  used  in  a  mill  may  be  divided  into  three 
general  classes  as  follows:  (1)  The  main,  or  head,  shaft, 
which  is  driven  directly  from  the  engine  or  other  source  of 
power;  this  shaft  is  sometimes  called  the  first,  or  prime, 
mover.  (2)  The  second  movers,  or  (as  they  are  gener¬ 
ally  known)  line  shafts;  these  are  the  main  driving  shafts 
of  each  room  and  derive  their  power  from  the  prime  mover. 
(3)  Countershafts  for  simply  transmitting  power  to  differ¬ 
ent  parts  of  the  room  or  for  making  changes  in  the  speed 
for  driving  some  particular  machine  or  machines;  these  are 
located  with  reference  to  the  positions  of  different  machines 
in  order  to  supply  them  with  power  as  economically  as  pos¬ 
sible.  Long  countershafts  are  classed  as  second  movers. 


SHAFT  COUPLINGS 

v 

3.  Where  a  long  line  of  shafting  is  required  to  economic¬ 
ally  distribute  the  power  it  becomes  necessary  to  couple  the 
several  sections  of  the  shaft  in  order  to  secure  the  transmis¬ 
sion  of  the  power.  Three  types  of  couplings  are  in  general 
use:  (1)  Fast,  or  permanent ,  coupIi?igs;  (2)  loose  couplings , 
or  clutches ,  by  means  of  which  shafts  may  be  connected  or 
disconnected  as  required;  (3)  friction  clutches,  which  are  loose 
couplings  that  hold  by  friction.  The  couplings  are  connected 
to  the  shafts  by  keys — tapered  pieces  of  metal  that  fit  into 
slots  in  both  the  shaft  and  the  coupling.  These  slots  are 
called  keyways,  or  key  seats. 


Fig.  1 

4.  Box,  or  muff,  couplings,  shown  in  Fig.  1,  consist 
of  a  cast-iron  cylinder  a  that  fits  over  the  ends  of  the  shafts; 
the  ends  are  prevented  from  moving  relative  to  each  other 
by  the  sunken  key  b.  The  keyway  is  cut  half  into  the  box 
and  half  into  the  shaft  ends;  quite  commonly  the  ends  of 


§7 


MECHANICAL  DEFINITIONS 


3 


the  shafts  are  enlarged  to  allow  the  keyway  to  be  cut  with¬ 
out  weakening  the  shafts. 

5.  A  clamp  coupling  is  shown  in  Fig.  2.  To  make 
this  coupling  the  face  of  each  half  for  the  joint  is  first 
planed  off,  the  holes  for  the  bolts  drilled,  and  the  two 
halves  bolted  together  with  pieces  of  paper  between  them, 
after  which  the  coupling  is  bored  out  to  the  exact  size  of 


Fig.  2 


the  shaft.  The  pieces  of  paper,  on  being  removed,  leave 
a  slight  space  between  the  halves,  and  the  coupling  when 
bolted  to  the  shaft  grips  it  firmly.  This  form  of  coupling  is 
very  easily  removed  or  put  on;  it  has  no  projecting  parts, 
and  may  be  used  as  a  driving  pulley,  if  desired.  The  key  in 
this  coupling  is  straight;  i.  e.,  not  tapered. 


Fig.  3 


6.  The  flange  coupling,  shown  in  Fig.  3,  consists  of 
cast-iron  flanges  a ,  a  keyed  to  the  ends  of  the  shafts.  To 
insure  a  perfect  joint,  the  flange  after  being  keyed  to  the 
shaft,  is  usually  faced  in  a  lathe.  The  two  flanges  are  then 


4 


MECHANICAL  DEFINITIONS 


7 


brought  face  to  face  and  bolted  together.  Sometimes  the 
ends  of  the  shafts  are  enlarged  to  allow  for  the  keyway. 

7.  Sometimes  a  clutch,  coupling,  shown  in  Fig.  4,  is 
used.  By  this  means  the  two  sections  of  the  shaft  may  be 
readily  disengaged;  so  that  if  any  part  of  the  room  is  shut 


Fig.  4 


down,  the  shafting  may  be  stopped  without  affecting  the  rest 
of  the  plant,  thus  saving  wear  on  the  equipment  and  a  great 
amount  of  power.  This  coupling  consists  of  a  piece  a  fast¬ 
ened  securely  to  the  end  of  one  shaft,  and  a  movable  piece  b 


Fig.  5 


sliding  on  a  key  let  into  the  end  of  the  other  shaft.  The 
two  halves  of  the  coupling  may  be  connected  or  discon¬ 
nected,  at  pleasure,  by  means  of  a  lever  having  a  yoke  at 
one  end  that  engages  with  the  groove  bt. 


§7 


MECHANICAL  DEFINITIONS 


5 


8.  Friction  couplings,  or  clutches,  are  used  as  loose, 
or  disengaging,  couplings  on  shafts  running  at  high  speeds; 
they  are  often  used  to  couple  wheels  or  pulleys  to  shafts. 
The  form  shown  in  Fig.  5  is  simple  in  construction  but  it  is 
difficult  to  put  'in  action;  besides,  it  is  necessary  to  exert 
a  pressure  in  a  horizontal  direction  in  order  to  keep  the 
clutch  in  action  and  this  causes  an  end  thrust  on  the  shaft. 
A  better  form  is  shown  in  Fig.  6.  The  shaft  n  carries  a 
flange  or  cylinder  a ,  while  the  shaft  m  has  a  ring  b  keyed  to 
it;  this  ring  is  split  and  fits  inside  the  flange  or  cylinder  a. 
The  split  ends  are  connected  by  a  screw  with  right-hand 
and  left-hand  threads;  the  screw  is  turned  by  the  lever  c, 


Fig.  6 


which  is  connected  by  the  link  d  to  the  sleeve  e.  When  the 
sleeve  is  pushed  toward  the  clutch,  the  rotation  of  the  screw 
throws  the  ends  /,  /  of  the  ring  b  apart,  and  thereby  causes 
the  ring  b  to  grip  the  flange  a  tightly.  A  clutch  of  this  form 
is  easy  to  operate,  and  produces  no  end  thrust  on  the  shaft. 

Friction  clutches  may  be  placed  in  contact  while  the  shaft 
is  running  at  high  speed,  but  to  do  this  with  the  ordinary 
form  of  clutch  coupling  would  break  the  coupling. 


SHAFT  HANGERS 

9.  For  convenience  and  safety,  the  shafting  is  generally 
supported  from  the  ceiling  by  hangers.  A  type  that  is 
often  used  in  textile-mill  equipments  is  shown  in  Fig.  7. 
This  hanger  consists  of  an  iron  frame  a  provided  with 
shoulders  through  which  holes  are  bored  so  that  it  can  be 


6 


MECHANICAL  DEFINITIONS 


§7 


securely  fastened  to  the  ceiling  by  means  of  lagscrews. 
This  frame  supports  a  box,  or  bearing,  b  in  which  the  shaft 
rotates,  and  below  which  there  is  a  drip  pan  c  that  catches 

the  oil  that  works  out  of  the  box. 
Though  all  hangers  are  not  pro¬ 
vided  with  drip  pans,  they  are  val¬ 
uable  adjuncts  to  them  in  weave 
rooms  where  the  oil  is  liable  to 
drop  on  fabrics  or  warps,  or  in 
other  places  where  cleanliness  is 
desired. 

The  box,  or  bearing,  of  a  hanger 
should  have  a  free  lateral  move¬ 
ment  and  also  a  vertical  rocking 
action,  in  order  that  it  may  con¬ 
form,  or  aline,  with  the  shaft. 
There  should  also  be  a  positive 
vertical  screw  adjustment,  in  order 
that  the  height  of  the  bearing  may  be  regulated  when  lining  up 
the  shafts.  The  length  of  the  box,  or  bearing  surface,  within 
which  the  shaft  turns,  varies  from  3|  to  5  times  the  diameter 
of  the  shaft,  according  to  the  ideas  of  different  makers.  It  is 
safe  to  use  one  from  4  to  5  times  the  diameter  of  the  shaft. 

The  inner  portion  of  the  box,  which  comes  in  contact  with 
the  shaft,  is  made  of  Babbitt  metal,  brass,  or  some  similar 
antifriction  metal  in  order  that  the  shaft  may  run  with  the 
least  possible  friction  and  heating. 

Shafting  may  be  easily  lined  up  by  adjusting  the  boxes  of 
the  hangers  by  means  of  screws  with  which  they  are  equipped. 
Shafting  should  be  lined  up  at  least  once  a  year  as  consider¬ 
able  damage  may  be  caused  by  imperfect  alinement.  Many 
mills  attend  to  this  every  6  months,  thereby  saving  wear  on 
the  boxes  of  the  hangers  and  guarding  against  the  liability 
of  a  broken  shaft. 

Hangers  should  preferably  be  attached  to  the  supporting 
beams  of  the  floor  above  and  as  near  to  a  line  of  posts  as 
possible  so  that  varying  weights  on  the  floor  above  will  not 
throw  the  shaft  out  of  line  to  any  great  extent. 


Fig.  7 


7 


MECHANICAL  DEFINITIONS 


7 


10.  Distance  Between  Hangers. — When  hangers  are 
put  up  they  should  be  lined  perfectly  true,  both  laterally  and 
vertically,  and  should  not  be  placed  too  far  apart.  The 
distance  between  the  bearings  should  not  be  great  enough 
to  permit  a  deflection  of  the  shaft  of  more  than  .01  inch  per 
foot  of  length.  Hence,  when  the  shaft  is  heavily  loaded  with 
pulleys,  the  bearings  must  be  closer  than  when  it  carries 
only  a  few.  Pulleys  that  transmit  a  large  amount  of  power 
should  be  placed  as  near  a  hanger  as  possible. 

The  following  table  gives  the  maximum  distances  between 
the  bearings  of  different  sizes  of  continuous  shafts  that  are 
used  for  the  transmission  of  power: 


Diameter  of  Shaft 
Inches 

Distance  Between  Bearings 

Feet 

Wrought-Iron  Shaft 

Steel  Shaft 

2 

I  I 

id 5 

3 

1 3 

13-75 

4 

15 

15-75 

5 

1 7 

18.25 

6 

19 

20.5 

7 

2  1  \ 

22.25 

8 

23 

24 

9 

25 

26 

11.  Types  of  Hangers. — Some  hangers  are  designed 
to  be  attached  to  posts  or  columns  instead  of  to  the  ceiling; 
in  this  case  they  are  called  post  hangers.  A  hanger  of 
this  type  is  shown  in  Fig.  8. 

Occasionally  supports  for  shafting  are  constructed  to  be 
attached  to  the  walls  of  the  mill  and  are  known  as  wall 
brackets.  A  common  type  of  wall  bracket  is  shown  in 
Fig.  9. 

When  the  support,  or  bearing,  of  a  shaft  is  designed  to 
rest  on  the  upper  side  of  a  horizontal  surface,  it  is  called  a 


8 


MECHANICAL  DEFINITIONS 


7 


.  Fig.  8  Fig.  9 


Fig.  10 


Fig.  11 


Fig. 12 


§7 


MECHANICAL  DEFINITIONS 


9 


pillow-block.  When  the  same  bearing  is  fastened  on  the 
under  side  of  a  horizontal  surface,  it  is  termed  an  inverted 
pillow-block.  A  pillow-block  bearing  is  shown  in  Fig.  10. 

When  it  is  necessary  to  pass  a  shaft  through  a  wall,  a 
special  bearing  known  as  a  wall  box  is  used,  a  view  of 
which  is  shown  in  Fig.  11.  This  bearing  is  designed  to  be 
built  into  the  brickwork  of  the  wall. 

A  floor  stand  is  a  support  for  a  shaft  running  near  the 
floor.  This  is  often  used  to  support  the  end  of  a  long 
shaft  that  projects  from  a  machine.  An  ordinary  type  of 
floor  stand  is  shown  in  Fig.  12. 


POWER  TRANSMISSION 


FRICTIONAL  CONTACT 

12.  Power  transmission  in  mill  work  is  accomplished 
by  different  methods  according  to  the  needs  of  the  individual 
mill;  but  where  a  great  amount  of  power  is  to  be  transmitted, 
it  generally  necessitates  the  employment  of  shafting,  pul¬ 
leys,  belts,  ropes,  gearing,  etc. 

Frictional  contact  is  sometimes  used  for  transmitting 
power  where  it  is  desired  to  apply  the  power  quickly  or  to 
disconnect  it  with  as  little  loss  of  time  as  possible.  The 
simplest  method  of  doing  this  is 
by  means  of  the  frictional  con¬ 
tact  between  the  circumferences 
of  circular  disks  or  pulleys,  as 
shown  in  Fig.  13.  Two  pul¬ 
leys  d,  /  are  shown  attached  to 
shafts  with  their  circumferences 
in  contact  at  a.  If  d  imparts 
motion  to  /,  then  d  is  the  driver  and  /  the  driven,  and  vice 
versa;  in  any  case  the  motion  that  is  imparted  is  in  the 
opposite  direction  to  that  of  the  driver. 

In  order  that  an  excessive  amount  of  slippage  may  not  take 
place,  the  face  of  the  driving  pulley  is  usually  made  of  wood 
or  is  covered  with  leather  or  paper,  while  the  driven  pulley 


10 


MECHANICAL  DEFINITIONS 


§7 


is  made  of  cast  iron;  if  any  degree  of  efficiency  is  to  be 
obtained  it  is  imperative  that  the  driving  and  driven  pulleys 
be  pressed  together  with  considerable  force.  The  amount 
of  power  that  may  be  transmitted  by  a  friction  drive  depends 
on  the  magnitude  of  this  force  together  with  the  coefficient 
of  friction  between  the  pulleys.  Frictional  contact  is  only 
feasible  when  small  amounts  of  power  are  to  be  transmitted. 
It  is  not  often  used  for  power  transmission  in  textile  mills. 


BELTING 

13.  Belts  running  over  pulleys  form  a  convenient  means 
for  transmitting  power,  but  they  are  not  suited  to  transmit  a 
precise  velocity  ratio,  owing  to  their  tendency  to  stretch  and 
to  slip.  For  driving  machinery,  however,  this  freedom  to 
stretch  and  to  slip  is  an  advantage,  since  it  prevents  shocks 
that  are  liable  to  occur  when  a  machine  is  suddenly 
started,  or  when  there  is  a  sudden  fluctuation  in  the  load. 
Three  kinds  of  belts  are  frequently  used  in  textile-mill  work: 
leather  belts,  cotton  belts,  and  rubber  belts. 

14.  Leather. — The  material  most  commonly  used  for 
belts  is  leather  tanned  from  ox  hides.  It  averages  -ft  inch 
in  thickness  and  is  obtained  in  strips  up  to  5  feet  long. 
Belts  of  any  required  length  are  made  by  joining  these  strips 
together.  The  wider  belts  are  made  from  the  thicker  por¬ 
tions  of  the  hide.  Leather  belts  should  be  oak-tanned,  as 
those  tanned  with  hemlock  bark  or  extracts  do  not  have  the 
same  strength  nor  do  they  last  as  long. 

15.  Cotton  helts  can  be  made  very  wide  and  without 
as  many  joints  as  leather  belts.  The  necessary  thickness 
is  obtained  either  by  sewing  together  from  four  to  ten  plies 
of  cotton  duck,  or  weaving  a  belt  as  a  ply  cloth  so  as  to  form 
a  fabric  of  several  plies  securely  interwoven  and  with  good 
selvages.  Cotton  belting  is  cheaper  than  leather. 

16.  Rubber  belts,  which  are  made  by  cementing  plies 
of  cotton  duck  with  india  rubber,  are  more  adhesive  than 
leather  belts  and,  therefore,  have  greater  driving  capacity. 


§7 


MECHANICAL  DEFINITIONS 


11 


They  are  largely  used  in  dye  houses,  or  other  places  where 
the  air  is  damp  or  full  of  steam,  as  under  these  conditions 
leather  will  quickly  stretch  and  rot. 

17.  Single  and  Double  Belts. — A  single  belt  is  one 
that  consists  of  only  one  thickness  of  hide.  A  double  belt 
consists  of  two  thicknesses  of  the  hide  cemented,  stitched,  or 
pegged  together  with  the  grain,  or  hair,  side  out.  They  are 
used  for  heavy  drives  where  a  large  amount  of  power  is 
transmitted.  All  belts  over  6  inches  in  width  should  be 
double.  The  ends  of  a  double  belt  should  be  tapered ,  or 
skived  down ,  and  then  cemented  together.  Double  belts  are 
always  made  of  leather  taken  from  the  center  of  the  hide 
along  the  backbone,  or  from  strips  taken  from  each  side  of 
the  center;  these  latter  make  the  best  belts  as  the  hide  is  apt 
to  be  thick  along  the  backbone.  The  strength  of  a  double 
belt  is  to  that  of  a  single  belt  as  10  is  to  7. 

Occasionally  belts  are  made  three-ply,  that  is,  they  consist 
of  three  thicknesses  of  hide. 

18.  Belt  Fastenings. — There  are  many  good  methods 
of  fastening  the  ends  of  belts  together,  but  lacing  is  gener¬ 
ally  used,  as  it  is  flexible  like  the  belt  itself,  and  runs  noise¬ 
lessly  over  the  pulleys.  The  ends  to  be  laced  should  be  cut 
squarely  across  and  the  holes 
in  each  end  for  the  lacings 
should  be  exactly  opposite 
each  other  when  the  ends  are 
brought  together.  Very  nar¬ 
row  belts,  or  belts  having  only 
a  small  amount  of  power  to 
transmit,  usually  have  only  one 
row  of  holes  punched  in  each 
end,  as  in  Fig.  14;  a  is  the  out¬ 
side  of  the  belt,  and  b  the  side  running  next  to  the  pulley. 
To  lace,  the  lacing  should  be  drawn  half  way  through  one 
of  the  middle  holes,  from  the  under  side,  as  for  instance 
through  1 ;  the  upper  end  should  then  be  passed  through  2, 
under  the  belt  and  up  through  3,  back  again  through  2  and  3, 


12 


MECHANICAL  DEFINITIONS 


§7 


through  4  and  up  through  5,  where  an  incision  is  made  in 
one  side  of  the  lacing,  forming  a  barb  that  will  prevent  the 
end  from  pulling  through.  The  other  side  of  the  belt  is 
laced  with  the  other  end,  it  first  passing  up  through  4. 
Unless  the  belt  is  very  narrow,  the  lacing  of  both  sides 
should  be  carried  on  at  once. 

Fig.  15  shows  a  method  of  lacing  where  double  lace  holes 
are  used,  b  being  the  side  to  run  next  to  the  pulley.  The 
lacing  for  the  left  side  is  begun  at  1,  and  continues  through 

2,  3,  4,  5 ,  6,  7,  6,  7,  4,  5,  etc.  A 
6-inch  belt  should  have  seven 
holes,  four  in  the  row  nearest 
the  end,  and  a  10-inch  belt,  nine 
holes.  The  edges  of  the  holes 
should  not  be  nearer  than  f  inch 
from  the  sides;  and  the  holes 
should  not  be  nearer  than 
1  inch  from  the  ends  of  the  belt. 
The  second  row  should  be  at 
least  If  inches  from  the  end. 
Another  method  is  to  begin  the  lacing  at  one  side  instead  of  in 
the  middle.  This  method  will  give  the  rows  of  lacing  on  the 
under  side  of  the  belt  the  same  thickness  all  the  way  across. 

19.  Although  a  laced  belt  with  the  holes  punched  as 
small  as  possible  is  the  best,  a  good  copper-riveted  belt  will 
give  good  service.  Belt  hooks  are  not  desirable  except  in 
cases  of  emergency  or  for  narrow  belts  or  those  that  have 
to  be  frequently  altered,  as  they  cut  the  belt  and  are  also 
liable  to  catch  in  the  clothing  of  the  operatives  and  thus 
cause  accidents. 

20.  Care  of  Belts. — 1.  Belts  should  be  run  with  the 
smooth,  or  grain,  side  next  to  the  pulley  for  the  following 
reasons:  ( a )  There  is  more  friction  of  the  belt  on  the  pul¬ 
ley  and,  therefore,  less  slipping  and  consequent  loss  of 
power,  {b)  The  center  of  strength  in  a  belt  is  located  one- 
third  of  the  distance  through  the  belt  from  the  flesh  side 
and  it  is  better  to  crimp  the  grain,  or  weak,  side  around  the 


§7 


MECHANICAL  DEFINITIONS 


13 


pulley  than  to  strain  it.  ( c )  The  stronger  side  of  the  belt 
receives  the  least  wear  when  run  in  this  manner. 

Some  authorities  recommend  that  the  flesh  side  of  a  belt 
be  run  next  to  the  pulleys,  and  while  this  is  contrary  to  the 
general  practice,  in  some  cases  it  gives  good  results. 

2.  The  lower  part  of  a  horizontal  or  inclined  belt  should 
be  the  driving  part;  then  the  slack  part  will  run  from 
the  top  of  the  driving  pulley.  The  sag  of  the  belt  will 
then  cause  it  to  encompass  a  greater  length  of  the  cir¬ 
cumference  of  both  pulleys.  Long  belts,  running  in  any 
direction  other  than  the  vertical,  work  better  than  short 
ones,  as  their  weight  holds  them  more  firmly  to  their 
work.  There  is,  however,  a  disadvantage  in  long  belts, 
since  they  greatly  increase  the  strain  on  the  bearings  of 
the  shaft. 

3.  The  accumulations  of  grease  and  gummy  matter  should 
be  frequently  removed  and  the  belts  dressed  with  castor  oil 
or  some  other  suitable  dressing  on  the  side  of  contact,  in 
order  to  keep  them  moist  and  pliable.  It  is  bad  practice  to 
use  rosin  to  prevent  slipping;  it  gums  the  belt,  causes  it 
to  crack,  and  prevents  slipping  for  only  a  short  time. 

4.  If  a  belt  properly  cared  for  persists  in  slipping,  a  wider 
belt  or  larger  pulleys  should  be  used;  the  latter  to  increase 
the  belt  speed.  Belts  should  not  be  run  tight,  as  the  strain 
thus  produced  will  wear  out  both  the  belt  and  the  bearings 
of  the  shaft. 


PULLEYS 

21.  Pulleys  are  wheels  that  are  provided  with  wide 
faces,  or  rims,  in  order  to  give  sufficient  frictional  contact 
with  the  belts  that  run  around  them  to  transmit  the  required 
amount  of  power.  They  are  designed  to  be  attached  to 
shafts  for  the  purpose  of  transmitting  their  power  or  for 
communicating  power  t®  them,  and  are  known,  respectively, 
as  driving  and  driven  pulleys. 

The  size  of  a  pulley  is  gauged  by  its  diameter,  the  width 
of  the  face,  and  the  bore,  or  diameter,  of  the  hole  designed 
for  the  reception  of  the  shaft.  Split  pulleys  are  made  with 


MECHANICAL  DEFINITIONS 


14 


§7 


a  large  bore,  the  bushings  (see  Art.  24)  being  made  to  suit 
any  diameter  of  shafting. 

22.  A  flat,  or  straiglit-faced,  pulley  is  one  whose 
face,  on  which  the  belt  runs,  is  parallel  to  the  shaft  to  which 
it  is  fastened.  A  section  of  one  is  shown  in  Fig.  16. 


- 


23.  A  crown,  or  crown-faced,  pulley  is  one  with  a 
greater  diameter  in  the  center  of  its  face  than  at  either  edge, 

a  section  of  its  rim  showing  a 
curve  on  "the  exterior  side,  as  seen 
in  Fig.  17. 

The  object  of  crowning  a  pulley 
is  to  keep  the  belt  moving  on  the 
center  of  its  face,  even  if  the  aline- 
ment  of  the  two  pulleys  on  which 
the  belt  runs  is  not  perfect.  The 
crown  of  the  pulley  accomplishes 
this  because  the  tendency  of  a 
running  belt  is  to  run  on  the  high¬ 
est  part  of  the  pulley,  which  in  this 
case  is  in  the  center  of  its  face. 

If  a  belt  is  placed  to  run  on  a 
conical  pulley,  it  tends  to  move 
up  to  the  largest  diameter.  To 
prevent  the  belt  from  running  off,  two  short,  conical  pulleys 
might  be  used,  placed  on  the  same  shaft  with  the  large  diam¬ 
eters  together.  From  this  the  crowned,  or  rounded,  form 
of  the  face  of  pulleys  has  been  developed. 

For  driving  shaft  pulleys,  the  crown  is  usually  from  tV  to 
h  inch  per  foot  of  width  for  high-speed  pulleys  and  i  inch 
per  foot  of  width  for  slow-speed  pulleys.  Machine  pulleys 
are  frequently  crowned  more  than  this,  sometimes  as  much 
as  4  inch  on  a  3-inch-face  pulley.  As  a  general  rule,  a  high¬ 
speed  pulley  requires  less  crown  than  a  slow-speed  pulley, 
and  a  pulley  of  large  diameter  less  than  one  of  small  diam¬ 
eter.  Pulleys  are  sometimes  crowned  by  drilling  holes 
through  their  faces  and  riveting  leather  laps  around  them 
with  copper  rivets. 


a»Mai 


Fig.  16 


Fig.  17 


§7 


MECHANICAL  DEFINITIONS 


15 


24.  Split  and  Solid  Pulleys. — Pulleys  are  either  split 
or  solid.  A  split  pulley  is  made  in  halves  and  is  attached 
to  the  shaft  by  clamping  the  hub  of  the  pulley  to  the  shaft  or 
to  segments  of  metal  or  wood.  These  segments  are  known 
as  a  bushing  and  are  inserted  between  the  shaft  and  the 
inside  of  the  hub. 

A  solid  pulley  is  attached  to  the  shaft  either  by  a  key  or 
by  a  setscrew,  which  is  threaded  in  the  hub  of  the  pulley  and, 
when  turned  in,  binds  the  hub  to  the  shaft.  Sometimes  two 
or  more  setscrews  are  used  for  fastening  a  solid  pulley  on 
a  shaft,  in  which  case  the  screws  are  placed  at  an  angle  of  90° 
to  one  another.  In  this  posi¬ 
tion  the  pulley  is  held  firmer 
than  with  the  screws  oppo¬ 
site  each  other  in  the  hub. 

Pulleys  may  be  made  of 
cast  iron,  wrought  iron,  or  of 
wood.  The  latter  are  not 
generally  constructed  entirely 
of  wood,  the  rim  only  being 
made  of  this  material;  it  is 
built  of  blocks  nailed  and 
glued  together.  Sometimes 
pulleys  are  made  of  wrought 
sheet  steel  or  iron,  the  pulley 
being  stamped  out  and  built 
up  with  rivets.  A  solid  iron  pulley  is  shown  in  Fig.  18;  it 
will  be  noticed  that  the  face  of  the  pulley  is  crowned.  This 
illustration  also  shows  the  method  of  placing  the  setscrews 
for  attaching  the  pulley  to  the  shaft.  The  hub  of  this  pulley 
is  key-seated,  so  that  it  may  be  attached  to  the  shaft  by 
means  of  a  key  if  desired. 

Solid  pulleys  are  placed  on  the  shaft  before  it  is  raised 
into  position  and  placed  in  the  hangers,  while  split  pulleys 
may  be  attached  or  removed  at  any  time  without  disturbing 
the  shaft. 

A  wood-rim  split  pulley  is  shown  in  Fig.  19;  it  will 
be  noticed  that  this  is  a  straight-faced  pulley. 


Fm.  18 


16 


MECHANICAL  DEFINITIONS 


7 


25.  A  drum  is  a  long,  cylindrical  pulley,  the  term  being 
generally  used  when  more  than  one  belt,  or  band,  is  driven 
by  the  same  pulley. 


Fig.  19  Fig.  20 

26.  A  slieave,  or  slieave  pulley,  is  one  with  a  groove 
or  grooves  cut  or  cast  in  its  face  for  the  reception  of  driving 
ropes,  or  bands.  Sheave  pulleys  are  sometimes  known  as 
grooved  pulleys.  An  ordinary  type  of  sheave  pulley  is  shown 
in  Fig.  20. 


Fig.  21 


27.  Flange  pulleys  are  often  used  on  various  machines 
and  are  sometimes  found  on  countershafts.  Different  types 
are  shown  in  Fig.  21,  the  flanges  being  designed  to  pre¬ 
vent  the  belt’s  running  off  or  coming  in  contact  with  another 
belt  running  on  the  same  pulley. 


7 


MECHANICAL  DEFINITIONS 


17 


28.  An  idle  pulley,  or  idler,  is  a  pulley  around  which 
a  belt  runs  without  imparting  motion  to  any  particular 
mechanism  other  than  the  pulley  itself,  which  is  usually 
loose  on  a  stud. 


29.  A  guide  pulley  is  an  idler  placed  in  such  a  position 
and  at  such  an  angle  that  it  guides  the  belt  on  to  a  pulley  on 
which  it  could  not  otherwise  run. 

30.  A  binder  pulley  is  an  idler  so  arranged  between 
driving  and  driven  pulleys  that  by  moving  it  in  a  certain 
direction  the  slack  of  the  belt  will  be  taken  up. 


a  f 


31.  Tight  and  Loose  Pulleys. — The  customary  method 
of  communicating  power  to  a  machine  is  to  have  the  machine 
connected,  by  means  of  a  belt,  with  a  pulley  on  a  driving 
shaft,  which  is  usually  located  near  the  ceil¬ 
ing,  but  sometimes  arranged  underneath  the 
floor,  the  driving  belt  passing  through  holes 
in  the  floor.  As  it  is  obviously  impossible 
to  control  the  machine  economically  by  means 
of  the  driving  shaft,  there  are  two  driven  pul¬ 
leys  on  the  machine,  one  of  which  is  fast  to  the 
shaft  of  the  machine  and  the  other  loose,  so 
that  it  may  rotate  without  imparting  motion 
to  the  machine.  Since  the  pulley  on  the  dri¬ 
ving  shaft  has  a  sufficient  width  of  face  to  admit 
of  shifting  the  belt,  it  will  be  seen  that  the 
machine  may  be  controlled  independently  of 
the  driving  shaft  by  shifting  the  belt  on  to  the 
tight  or  the  loose  pulley,  according  to  whether 
it  is  desired  to  start  or  stop  the  machine. 

Such  an  arrangement  is  spoken  of  as  tight 
and  loose  pulleys  and  is  shown  in  Fig.  22. 

In  this  illustration  the  overhead  driving 
shaft  is  marked  a  and  has  the  driving  pulley  b 
attached  to  it.  This  pulley  has  a  face  of  sufficient  width  so 
that  the  belt  may  be  moved  sidewise  a  trifle  more  than  its 
own  width.  The  motion  may  be  transmitted  by  the  belt  from 
the  driving  pulley  b  to  the  loose  pulley  c,  in  which  case  the 


Pig.  22 


18 


MECHANICAL  DEFINITIONS 


§7 


machine  will  not  be  driven,  as  the  loose  pulley  will  simply 
turn  on  the  driving  shaft  e  of  the  machine.  If,  however,  the 
belt  is  shifted  so  that  the  power  is  transmitted  from  the  dri¬ 
ving  pulley  b  to  the  tight  pulley  d,  which  is  keyed  to  the  shaft  e 
with  the  key  /,  motion  will  be  imparted  to  the  machine. 

32.  The  belt  is  moved,  or  shipped,  from  the  loose  to  the 
tight  pulley,  or  vice  versa,  either  by  hand  or  by  means  of  a 
belt  shipper.  The  usual  type  of  belt  shipper  consists  of  a 


rod  moving  in  guides  in  a  direction  parallel  to  the  shaft  that 
carries  the  tight  and  loose  pulleys,  and  provided  with  a  fork, 
between  the  tines  of  which  the  belt  runs.  When  the  rod  is 
moved  in  the  direction  of  the  loose  pulley,  the  fork  pressing  on 
the  side  of  the  belt  ships  it  on  to  that  pulley,  and  vice  versa. 

33.  Speed  cones,  or  step  pulleys,  are  used  in  machines 
where  different  speeds  are  required;  this  variation  in  speed 
is  obtained  by  means  of  pulleys  of  different  diameters.  The 


7 


MECHANICAL  DEFINITIONS 


19 


usual  method  of  making  step  pulleys,  or  speed  cones,  is  to 
have  pulleys  of  different  diameters  cast  in  one  piece  or  built 
up  with  wood  so  that  a  series  of  faces  is  formed  for  the 
belt  to  run  on.  A  similar  pulley  is  used  for  the  driven 
pulley,  but  is  placed  on  the  shaft  in  a  reversed  position,  so 
that  when  the  belt  is  on  the  largest  diameter  of  one  pulley, 
it  is  on  the  smallest  diameter  of  the  other  pulley,  and  vice 
versa.  The  diameters  of  the  intermediate  steps  are  of  course 
proportionally  changed  because  the  sum  of  the  diameters  of 
any  two  opposite  steps  must  be  constant  so  that  the  belt  will 
be  of  the  right  length  for  any  position  and  will  have  the 
same  tension  when  on  any  pair  of  pulleys. 

Fig.  23  shows  an  arrangement  of  speed  cones  that  is  very 
often  adopted.  In  this  case  the  driving  speed  cone  e  is  placed 
on  a  countershaft,  which  is  also  provided  with  a  tight  pulleys 
and  a  loose  pulley  d,  by  means  of  which  its  motion  may  be 
controlled  independently  of  the  main  shaft  of  the  room  from 
which  it  is  driven.  A  belt  k  connects  the  driving  speed  cone 
with  the  driven  cone  /,  which  is  fastened  to  the  shaft  h  of  the 
machine.  The  belt  may  be  shifted  from  one  step  of  the 
driving  and  driven  cones  to  another  in  order  to  secure  differ¬ 
ent  diameters  of  driving  and  driven  pulleys,  thus  producing 
a  wide  variation  of  possible  speeds  of  the  shaft  h. 

With  the  belt  on  the  largest  diameter  of  the  driving  cone  e 
and  the  smallest  diameter  of  the  driven  cone  /  the  maximum 
speed  of  the  machine  is  attained.  The  minimum  speed  of 
the  machine  is  produced  with  the  belt  on  the  smallest  diameter 
of  e  and  the  largest  diameter  of  /.  With  the  belt  in  interme¬ 
diate  positions,  speeds  ranging  between  the  maximum  and 
minimum  are  obtained.  It  should  be  noted  that,  when  speak¬ 
ing  of  a  cone  pulley  as  having  a  certain  number  of  steps,  the 
number  of  steps  is  one  less  than  the  number  of  pulleys  on  the 
cone.  Thus,  the  cones  in  Fig.  23  are  four-step  cones  and 
have  five  pulleys.  Consequently,  if  a  cone  pulley  (or  cone) 
is  spoken  of  as  having  five  steps,  there  are  six  pulleys  and 
six  changes  of  speed. 

In  Fig.  23,  the  tight  and  loose  pulleys  are  on  the  counter¬ 
shaft  instead  of  the  driving  shaft  of  the  machine.  The 


20 


MECHANICAL  DEFINITIONS 


7 


machine  is  stopped  or  started  by  moving  the  lever  n}  which 
operates  the  bar  in.  Attached  to  m  is  the  fork  /  between  the 
tines  of  which  the  countershaft  belt  passes.  By  moving  the 
lever  to  the  left  the  belt  is  shipped  ofi  to  the  tight  pulley  c, 
thus  imparting  motion  to  the  countershaft  and  then  to  the 
cones  and  the  machine  driven  by  them.  Moving  the  lever  to 
the  right  ships  the  belt  from  c  on  to  the  loose  pulley  d,  thus 
stopping  the  machine. 


34.  From  step  pulleys,  or  speed  cones,  the  tapered,  or 
true,  cones,  which  have  no  steps  for  the  belt  to  rest  on, 
have  originated,  the  belt  being  automatically  shifted  along 
their  surfaces  by  a  belt  guide,  thus  securing  the  gradual 
variation  in  speed  that  is  necessary  in  certain  classes  of 
textile  machinery. 

35.  Dii*ection  of  Rotation. — The  connection  between 
two  pulleys  by  a  belt  or  rope  can  be  made  in  two  ways; 
namely,  by  an  open  belt,  as  shown  in  Fig.  24,  or  by 
a  crossed  belt,  Fig.  25.  When  a  pulley  is  driven  by  an 


open  belt,  its  direction  of  rotation  is  the  same  as  that  of  the 
driving  pulley,  as  indicated  by  the  arrows  in  Fig.  24.  When 
a  pulley  is  driven  by  a  crossed  belt,  its  direction  of  rotation 


§7 


MECHANICAL  DEFINITIONS 


21 


is  opposite  to  that  of  the  driving  pulley,  as  shown  by  the 
arrows  in  Fig.  25. 

36.  Countershafts. — In  practice  there  is  a  limit  to  the 
size  of  pulleys  that  can  be  used,  and  therefore  to  the  ratio 
of  speeds  that  can  be  obtained  from  two  shafts.  The  size  of 
driving  pulleys  is  limited  by  the  distance  between  the  ceiling 
and  the  shaft  and  also  by  the  size  of  the  driven  pulley.  For 
instance,  an  extremely  large  pulley  cannot  be  used  to  drive 
an  extremely  small  one  to  obtain  a  desired  speed,  because 
the  belt  will  slip  on  the  smaller  pulley  owing  to  the  lack  of 
sufficient  surface  contact. 

If  the  belt  is  applied  with 
sufficient  tension  to  over¬ 
come  this,  an  undue  strain 
will  be  placed  on  the  bear¬ 
ings  of  the  shaft  and  they 
will  become  heated.  To 
overcome  this  difficulty  an 
additional  shaft,  called  a 
countershaft,  is  used  and. 
in  some  cases,  more  than 
one  is  necessary  before  the 
desired  speed  can  be  ob¬ 
tained  with  an  economical 
transmission  of  power. 

The  transmission  of 
power  in  this  manner  re¬ 
quires  one  pulley  on  the 
first,  or  driving,  shaft  and  one  on  the  driven  shaft,  while 
each  intermediate  shaft,  or  countershaft,  requires  two  pul¬ 
leys  of  different  diameters.  An  arrangement  of  this  kind 
is  shown  in  Fig.  26.  If  a  is  the  driving  pulley  and  d  the 
driven,  the  speed  of  d  is  greater  than  that  of  a\  but  if  d  is 
the  driver  and  a  the  driven,  the  speed  of  a  is  less  than  that 
of  d.  If  pulleys  b  and  c  are  of  the  same  size,  the  power  will 
be  transmitted  through  the  countershaft  without  any  change 
of  speed  being  given  by  it  to  d. 


22 


MECHANICAL  DEFINITIONS 


§  7 


37.  Speed  of  Pulleys. — Pulleys  over  4  feet  in  diameter 
and  flywheels,  especially  cast-iron  ones,  should  never  be 
speeded  so  fast  that  their  surface  velocity  exceeds  5,000  feet 
per  minute,  since  there  will  be  a  danger  of  their  bursting. 
Many  authorities  give  3,750  feet  per  minute  as  a  limit  to 
the  surface  speed  of  large  pulleys.  Smaller  pulleys  may 
have  a  higher  surface  velocity,  but  excesses  should  be 
avoided. 


NON-PARALLEL  SHAFTS 


c 


J3 


U 


38.  In  the  drives  previously  mentioned  the  driving  and 
driven  shafts  have  been  parallel  to  each  other.  But  condi¬ 
tions  that  frequently  arise  where  power  is  required  to  be 

transmitted  by  belts  or  ropes  are: 

(1)  when  the  driving  and  driven 
shafts  are  at  right  angles  to  each 
other  and  are  not  in  the  same  plane; 

(2)  when  the  driving  and  driven 
shafts  are  non-parallel  and  are  both 
in  the  same  plane. 

39.  Quarter-Turn. — When 
shafts  are  arranged  as  in  the  first 
instance,  the  pulleys  must  be  so 
placed  that  the  belt  is  delivered  from 
one  pulley  into  a  plane  passing 
through  the  center  of  the  face  of 
the  other  pulley.  The  arrangement 
shown  in  Fig.  27  is  known  as  a 
quarter- turn  because  the  belt  is 
given  a  quarter  twist.  A  connec¬ 
tion  of  this  kind  can  only  be  driven 
in  the  direction  indicated  by  the 
If  the  direction  of  the  belt  is  reversed 
it  will  run  off  the  pulleys  unless  a  guide  pulley  is  used. 

The  easiest  and  most  convenient  way  of  fixing  the  posi¬ 
tion  of  quarter-turn  pulleys  is  to  plumb  the  leaving  sides  of 
each  pulley;  that  is,  drop  a  plumb-line  from  the  center  of  the 


Fig.  27 


arrows  on  the  belt. 


§7 


MECHANICAL  DEFINITIONS 


23 


face  of  the  leaving  side,  where  the  belt  leaves  the  driving 
pulley,  and  arrange  the  driven  pulley  so  that  the  plumb- 
line  shall  just  touch  the  center  of  its  face  on  the  side  from 
which  the  belt  leaves  it.  This  is  shown  by  the  two  pul¬ 
leys  at  the  top  of  Fig.  27,  which  represents  a  plan  of  two 
quarter-turn  pulleys  as  seen  from  above. 

The  objection  to  a  quarter-turn  belt  is  that,  when  the  angle 
at  which  the  belt  is  drawn  off  the  pulleys  is  large,  the  belt 
is  strained,  especially  at 
the  edges,  and  it  does 
not  hug  the  pulleys  well. 

Small  pulleys  placed 
quite  a  distance  apart, 
with  narrow  belts  give 
the  best  results,  from 
which  it  follows  that 
quarter-turn  belts  are 
not  well  suited  to  trans¬ 
mit  much  power. 

Fig.  28  shows  how 
the  arrangement  can  be 
improved  by  placing  a 
guide  pulley  against 
the  loose  side  of  the 
belt.  The  driver  d  re¬ 
volves  in  a  left-hand 
direction,  making  a  b  the 
driving,  or  tight,  side  of  the  belt.  To  determine  the  posi¬ 
tion  of  the  guide  pulley,  select  some  point  in  the  line  a  b, 
as  g.  When  the  pulleys  differ  in  diameters  this  point  should 
be  somewhat  nearer  the  smaller  pulley.  Draw  lines  eg  and 
eg;  the  middle  plane  of  the  guide  pulley  should  then  pass 
through  the  two  lines.  Looked  at  from  a  direction  at  right 
angles  to  pulley  /,  line  eg  coincides  with  a  b;  looking  at 
right  angles  to  pulley  d,  line  eg  also  coincides  with  a  b. 


Fig.  28 


40.  If  two  shafts  are  not  parallel  to  each  other  and  are 
both  in  the  same  plane,  they  may  be  driven  by  a  belt  by 


24 


MECHANICAL  DEFINITIONS 


§7 


means  of  guide  pulleys  as  shown  in  Fig.  29.  In  a  case  like 
this  the  guide  pulleys  must  be  placed  so  that  the  active  part 
of  their  surfaces  will  be  tangent  to  planes  passing  through 
the  centers  of  the  faces  of  the  driving  and  driven  pulleys  at 
the  intersection  of  these  planes,  as  shown  in  the  upper,  or 
plan,  view  in  Fig.  29.  They  must  also  be  placed  so  that 


their  center  planes  are  tangent  to  the  surfaces  of  the  dri¬ 
ving  and  driven  pulleys,  as  shown  in  the  elevation,  or  lower, 
view.  A  pair  of  guide  pulleys  and  their  shaft,  as  illustrated 
in  Fig.  29,  are  known  as  a  mule  pulley  stand. 


ROPE  TRANSMISSION 

41.  Many  American  mills  are  introducing  rope  drives 
for  transmitting  power,  especially  for  the  main  drives  from 
the  engine,  for  which  this  method  is  particularly  adapted. 
The  distance  to  which  power  can  be  transmitted  by  means  of 
sheave  pulleys  and  ropes  is  practically  unlimited,  as  is  also 
the  amount  of  power.  Except  for  very  short  distances,  rope 
driving  is  the  cheapest  method  of  transmitting  power,  being 
economical  not  only  in  the  first  cost,  but  in  the  maintenance. 
This  in  itself  is  an  important  item.  An  evenness  of  motion 
that  cannot  •  be  obtained  by  any  other  system  of  power 


§7  MECHANICAL  DEFINITIONS  25 

transmission  is  obtained  by  transmitting  power  in  this  man¬ 
ner;  this  is  due  to  the  lightness,  elasticity,  and  slackness  of  the 
rope,  which  takes  up  all  inequalities  between  the  power  and 
the  load.  Rope  drives  are  noiseless  because  of  the  flexibility 
and  lubrication  of  the  rope  and  because  of  the  air  passage 
underneath  the  rope,  owing  to  the  Y-shaped  groove  in  which 
it  runs.  An  exact  alinement  of  the  driving  and  driven 
pulleys  is  not  necessary  when  ropes  are  used,  and  by  properly 
placing  idle  pulleys  power  may  be  transmitted  in  any  desired 


direction.  The  security  that  a  rope  drive  affords  against 
shut-downs  due  to  the  crippling  of  the  drive  is  one  of  the 
great  advantages  of  this  system.  This  is  due  to  the  fact 
that  before  breaking,  the  rope  stretches  excessively,  though 
gradually,  thus  giving  warning  that  it  should  be  replaced. 
The  absence  of  electrical  disturbances  and  the  almost  total 
immunity  from  slip  are  among  the  many  advantages  that 
may  well  be  claimed  by  this  system  for  power  transmission. 

42.  There  are  two  systems  of  rope  transmission  in 
common  use.  In  the  first,  the  transmission  is  effected  by 
several  parallel,  independent  ropes  that  pass  around  the 


26 


MECHANICAL  DEFINITIONS 


flywheel  of  the  engine  and  the  pulley  or  pulleys  to  be  driven. 
Each  rope  is  made  quite  taut  at  first,  but  stretches  until  it 
slips,  after  which  it  is  respliced.  A  good  example  of  this 
system  is  shown  in  outline  in  Fig.  30.  The  flywheel  m 
carries  thirty-five  parallel  ropes  that  distribute  power  to  the 
pulleys  a ,  b,  c,  d ,  e,  /,  located  on  the  five  floors  of  the  mill. 
The  ropes  are  distributed  as  follows:  a,  four  ropes;  b  and  c, 
five  ropes  each;  d,  e,  and  /,  seven  ropes  each.  A  secondary 
system  of  ropes  drives  the  pulleys^,  h,  k ,  /. 


In  the  second  system  of  rope  transmission,  a  single  rope 
is  carried  around  the  pulleys  as  many  times  as  is  necessary, 
to  transmit  the  required  power;  the  necessary  tension  is 
obtained  by  passing  a  loop  of  the  rope  around  a  weighted 
pulley.  An  example  of  this  system  is  shown  in  Fig.  31. 
The  rope  is  wrapped  continuously  around  the  flywheel  d  and 
the  driven  pulley  e.  From  the  last  groove  of  e ,  the  rope  is 
led  over  the  idlers  f,g,  which  are  set  at  such  an  angle  as  to 
lead  it  back  to  the  first  groove  in  d.  The  weight  w  is 
attached  to  the  pulley  /,  which  is  movable  along  the  rod  Ji. 
The  movement  of  the  pulley  /,  therefore,  takes  up  the 
stretch  of  the  rope,  and  keeps  it  always  at  the  same  tension. 
Pulleys  may  be  attached  to  the  shaft  of  the  pulley  e ,  and  the 


§7 


MECHANICAL  DEFINITIONS 


27 


power  received  by  e  may  thus  be  transmitted  to  any  desired 
points. 

The  first  of  the  above  systems  of  transmission  is  used 
chiefly  in  Europe;  the  second,  in  the  United  States.  The 
ropes  generally  used  are  of  Manila  hemp  or  cotton,  some¬ 
times  with  a  wire  core.  For  transmitting  power  long  dis¬ 
tances,  especially  where  the  rope  is  exposed  to  the  weather, 
a  wire  rope  is  used.  For  inside  drives  the  cotton  rope 
without  a  wire  core  is  suitable. 

43.  Next  in  importance  to  the  rope  are  the  grooved 
pulleys,  or  sheaves,  on  which  the  rope  runs.  The  grooves 
are  made  of  metal  or  wood  and  must  be  smooth,  in  order  to 
prevent  the  rope  from  wearing,  and  true,  to  keep  it  from 
swaying.  These  grooves  are  made  Y-shaped  so  that  they 
may  grip,  or  bind,  the  rope  and  not  allow  it  to  slip;  the 
best  forms  are  those  in  which  the  sides  approach  each 
other  at  an  angle  of  from  45°  to  60°.  The  rope  does  not 
touch  the  bottom  of  the  groove,  but  is  wedged  in  between 
the  sides. 

An  idle  pulley  for  a  rope  drive  may  have  half-round 
grooves,  as  there  is  no  necessity  for  gripping  the  rope  at 
this  point. 


GEARING 

44.  Gearing. — The  transmission  of  power  for  short 
distances  at  slow  speeds,  as  between  the  driving  and  driven 
shaft  of  a  machine,  is  generally  accomplished  by  means  of 
gears.  A  gear  is  a  disk  with  a  hole  in  its  center,  so  that  it 
can  be  readily  attached  to  a  shaft.  Its  face  is  cut,  or  cast, 
with  teeth  that  are  so  designed  as  to  mesh  with  the  teeth 
of  another  gear,  thus  producing  motion  by  a  direct  pressure 
of  the  teeth  of  one  gear  on  the  teeth  of  the  other. 

Gearing  is  ordinarily  made  of  cast  iron;  if  great  strength 
is  required,  steel  may  be  used.  Gears  that  are  called  on 
to  resist  shocks  may  be  made  of  gun  metal  or  phosphor 
bronze.  Fast-running  gears  are  sometimes  made  with 
wooden  cogs,  or  of  rawhide  or  fiber,  instead  of  metal. 


28 


MECHANICAL  DEFINITIONS 


§7 


45.  Teetli  is  the  name  applied  to  the  projections  around 
the  circumference  of  a  gear  when  they  are  made  from  the 
same  piece  as  the  body  of  the  gear;  while  cogs  is  the  name 
used  when  they  are  made  of  separate  material  and  fastened 
to  the  rim  of  the  gear.  Gears  with  cogs  inserted  in  their  rims 
are  called  mortise  gears.  Toothed  gears  are  in  gear  when 
their  teeth  are  engaged,  and  out  of  gear  when  separated. 

Fig.  32  shows  two  gears,  either  of  which  may  be  the  driver 
and  communicate  motion  to  the  driven  gear  by  direct  contact 


of  its  teeth.  It  will  be  noticed  that  the  direction  in  which 
the  driven  gear  rotates  is  opposite  to  that  of  the  driver. 

46.  The  pitch  circle  of  a  gear  is  an  imaginary  circle 
described,  with  the  axis  of  rotation  for  a  center,  through  the 
point  of  contact  of  the  teeth  of  one  gear  with  those  of  another 
gear.  It  is  its  effective  circumference  and  determines  its  ratio 
of  velocity  when  working  with  other  gears.  In  Fig.  32  the 
dot-and-dash  lines  represent  the  pitch  circles  of  the  two  gears. 

Suppose  that  two  shafts  10  inches  apart  are  to  be  connected 
by  gears  and  are  required  to  revolve  in  the  ratio  of  3  to  2; 
then  the  diameters  of  the  pitch  circles  of  the  gears  will  be  8 


7 


MECHANICAL  DEFINITIONS 


29 


and  12  inches.  Two  circles  centered  on  the  shafts  and  having- 
radii  of  4  and  6  inches  will  give  the  required  ratio  of  speeds 
by  rolling  contact;  thus,  this  regulates  the  diameters  of  the 
pitch  circles  of  two  gears  to  give  the  required  relative 
velocity. 

47.  The  circular  pitch  of  a  gear  is  the  distance 
between  the  centers  of  two  consecutive  teeth  measured  on 
the  pitch  circle  or,  as  it  is  sometimes  called,  the  pitch  line. 

48.  The  diametral  pitch  of  a  gear  is  equal  to  the 
number  of  teeth  on  its  circumference  divided  by  the  number 
of  inches  in  the  diameter  of  the  pitch  circle.  In  order  to 
mesh  and  run  together,  gears  must  be  of  the  same  pitch. 

49.  The  part  of  the  gear-tooth  outside  of  the  pitch  circle 
is  called  the  point,  or  addendum,  of  the  tooth  and  the 
part  inside  the  pitch  circle  is  called  the  root  of  the  tooth. 

The  root  of  a  tooth  is  usually  about  four-sevenths,  and  the 
point  about  three-sevenths,  the  length  of  the  tooth.  This 
allows  the  point  of  the  tooth  to  clear  the  bottom  of  the  space 
between  the  teeth  of  the  other  gear  by  one-seventh  of 
its  length. 

50.  The  amount  of  space  by  which  the  width  between 
the  teeth  is  greater  than  the  thickness  of  the  tooth  is  called 
the  backlasli.  The  dimensions  of  gears  are  given  by  the 
diameter  of  the  bore,  by  the  number  of  teeth  that  the  gear 
contains,  and  by  the  pitch. 

51.  Trains  of  Gears. — When  the  driving  and  driven 
shafts  are  so  far  apart  as  to  necessitate  the  use  of  gears  of 
too  large  diameters,  or  when  the  desired  ratio  of  velocity  is 
such  that  one  gear  must  have  an  excessive  diameter,  it  is 
necessary  to  interpose  gears  between  the  driving  and  driven 
shafts.  When  gears  are  combined  in  this  manner  they  are 
known  as  a  train  of  gears.  Fig.  33  shows  a  train  of  gears 
thus  arranged,  the  motion  being  transferred  from  the  shaft  e 
to  the  shaft  /  by  means  of  the  gears  a ,  d ,  c,  b.  The  gears  d ,  c 
are  used  to  transfer  the  motion  of  a  to  b  and  to  give  the 
latter  a  velocity  different  from  that  of  the  gear  a. 


30 


MECHANICAL  DEFINITIONS 


§7 


52.  Spur  gears  have  the  teeth  arranged  on  the  outer 
rim  parallel  to  the  shaft  on  which  the  gear  is  placed.  The 
action  of  a  pair  of  these  gears  may  be  regarded  as  that  of 


two  rolling  cylinders  minus  any  slip.  This  type  of  gear  is 
one  that  is  largely  used  for  transmitting  power  between 
parallel  shafts  and  is  shown  in  Fig.  33. 


Fig.  34 


53.  Bevel  gears  are  designed  to  transmit  power  from 
one  shaft  to  another  when  the  shafts  are  in  the  same  plane 
but  not  parallel.  The  action  of  a  pair  of  bevel  gears  may  be 
regarded  as  that  of  two  rolling  cones  whose  axes  intersect 


§7 


MECHANICAL  DEFINITIONS 


31 


at  their  apexes.  Fig.  34  shows  an  ordinary  type  of  bevel 
gears.  When  two  bevel  gears  of  the  same  size  are  designed 
for  shafts  that  are  at  right  angles  to  each  other,  as  in  Fig.  34, 
they  are  also  known  as  miter  gears. 


54.  A  ratchet  gear,  as  shown  in  Fig.  35,  is  designed  to 
engage  with  a  small  piece  of  metal  called  a  pawl.  The 
teeth  are  so  shaped  that  motion  can  be  imparted  to  the  gear 
in  one  direction  only  by  a  pawl  similar  to  b.  It  is  held  from 


turning  in  the  other  direction 
by  the  action  of  the  pawl  b, 
which  engages  with  the 


Fig.  36 


teeth.  Owing  to  the  inclination  of  the  teeth,  when  the  ratchet 
a  is  turning  in  the  proper  direction,  the  pawl  slips  over  the 
backs  of  the  teeth. 

Sometimes  a  ratchet  gear  is  driven  a  definite  number  of 
teeth  at  a  time  by  the  action  of  an  oscillating  pawl,  and  is 
then  held  by  a  stationary  pawl  while  the  driving  pawl  is  work¬ 
ing  backwards.  At  other  times  the  ratchet  is  attached  to  a 
shaft  having  an  intermittent  motion,  the  pawl  preventing  the 
shaft  from  turning  backwards  during  the  periods  of  rest,  but 
offering  no  resistance  to  its  forward  motion. 

55.  Sprocket  gears,  or  wheels,  are  designed  to  be 
driven  by  and  to  drive  chains.  A  common  type  of  this  gear 
is  shown  in  Fig.  36. 


32 


MECHANICAL  DEFINITIONS' 


§7 


56.  Spiral  gears  are  those  that  have  the  teeth  cut 
obliquely  on  their  surfaces  and  are  used  to  connect  shafts 
that  form  an  angle  with  each  other. 

57.  Star  gears  are  those  having  recesses  cut  in  them  at 
wide  intervals  and  are  driven  by  special  contrivances,  often 
by  a  pin.  This  pin  is  often  attached  to  a  disk  and  engages 


with  the  star  gear  at  each  rotation  of  the  shaft.  A  star  gear 
is  always  a  driven  gear.  * 

58.  Mangle  gears  are  used  to  reverse  their  own  direc¬ 
tion  of  rotation  and  are  always  driven  gears.  They  are  either 
eccentric  or  concentric.  When  concentric  the  center  of  the 
pitch  circle  coincides  with  the  axis  of  rotation,  and  when 
eccentric  the  center  of  the  pitch  circle  is  removed  from  the 
axis  of  rotation. 


§7 


MECHANICAL  DEFINITIONS 


33 


An  eccentric  mangle  gear  /  is  shown  in  Fig.  37.  The  gear 
is  of  somewhat  peculiar  construction,  having  instead  of  the 
ordinary  gear-teeth,  a  number  of  pegs,  or  pins,  fastened  in 
its  rim.  These  are  designed  to  mesh  with  the  driving  pinion  e 
fastened  to  the  shaft  d,  by  means  of  which  motion  is  imparted 
to  the  mangle.  The  shaft  d  is  carried  in  a  wide  guide,  or 
bearing,  dx  so  that  the  pinion  e  may  accommodate  itself  to 
the  eccentricity  of  the  mangle  and  engage  with  either  the 
outside  or  the  inside  of  the  row  of  pins,  thus  enabling  the 
mangle  to  be  driven  in  either  direction.  The  shaft  d  projects 
beyond  the  pinion  e  a  slight  distance  and  engages  with  a 
flange  A  extending  entirely  around  the  gear  on  the  inside  of 
the  pins  and  part  way  around  the  outside.  By  this  means 
the  pinion  is  guided  from  the  outside  to  the  inside  of  the 
row  of  pins,  and  vice  versa. 

Suppose  that  in  Fig.  37  the  shaft  d  is  rotating  in  the  direc¬ 
tion  shown  by  the  arrow;  then  the  pinion  e  will  work  around 
on  the  inside  of  the  pins  on  the  mangle  until  the  last  pin  A 
is  reached,  when  the  flange  A  will  guide  the  gear  to  the  out¬ 
side  of  the  row  of  pins  and  the  direction  of  the  rotation  of 
the  mangle  will  be  reversed.  This  motion  will  continue 
until  the  other  end  of  the  row  of  pins  is  reached,  whereupon 
the  flange  will  guide  the  pinion  back  to  the  inside  of  the  row 
of  pins  and  the  direction  of  rotation  of  the  mangle  will  again 
be  reversed.  It  is  not  necessary  to  have  the  outside  flange 
pass  entirely  round  the  mangle  gear,  since  when  the  teeth 
are  engaging  with  the  outside  of  the  pins  on  the  portion  of 
the  gear  that  is  farthest  from  its  axis,  the  pinion  is  held  in 
contact  with  the  teeth  of  the  mangle  by  means  of  the  shaft  d, 
which  in  this  case  is  pressed  against  the  end  of  the  wide 
bearing  dx. 

A  mangle  gear  is  said  to  perform  a  complete  cycle  of  its 
movements  in  making  a  double  revolution;  that  is,  one  revo¬ 
lution  in  one  direction  and  one  in  the  opposite  direction. 

59.  A  worm  is  an  endless  screw  designed  to  mesh  with 
a  gear  called  a  worm-gear.  A  worm  and  a  worm-gear  are 
shown  in  Fig.  38.  Worms  are  single-threaded  when  they 


34 


MECHANICAL  DEFINITIONS 


§7 


have  a  single  thread  cut  on  them  and  double-threaded  when 
they  have  two  threads.  A  worm  is  always  a  driver,  a  single- 
threaded  worm  driving  the  worm-gear  one  tooth  for  each 
revolution  and  a  double-threaded  worm  moving  it  two  teeth. 

Occasionally  worms  are  met  with  that  have  three  threads 
cut  on  them.  Worms  furnish  a  means  of  reducing  a  great 
velocity  of  a  shaft  to  the  slow  speed  of  the  worm-gear.  In 

calculations  a  worm  is  counted 
as  a  one-tooth  gear  if  single- 
threaded  and  as  a  two-tooth 
gear  if  double-threaded. 

60.  An  annular  gear  is 
one  whose  teeth  are  cut  on  the 
inside  of  its  rim,  or  circum¬ 
ference. 

61.  A  rack  is  a  straight  bar 
with  teeth  cut  on  its  surface. 

62.  A  pinion  is  a  small 
gear  and  is  in  most  cases  a  spur 
gear. 

63.  Eccentric  gears  are 
gears  that  rotate  about  points 

not  centrally  located  with  regard  to  their  pitch  circles.  A 
pair  of  eccentric  gears  require  setting  in  the  proper  posi¬ 
tion  before  they  will  run;  otherwise,  they  will  bind  and  spring 
the  shafts  to  which  they  are  attached. 

A  pair  of  eccentric  gears  is  always  marked — in  some  cases, 
by  nicking  with  a  cold  chisel  two  teeth  on  one  gear  and  one 
tooth  on  the  other.  When  setting  the  gears,  place  them 
together  so  that  the  single  nicked  tooth  on  one  gear  will 
mesh  between  the  two  nicked  teeth  on  the  other  gear. 

64.  Elliptic  gears  are  made  in  the  form  of  an  ellipse 
and  are  marked  and  set  after  the  manner  of  eccentric  gears. 
Elliptic  and  eccentric  gears  are  used  for  producing  a  variable 
speed  of  the  driven  shaft  during  each  rotation  of  the  pair 
of  gears. 


7 


MECHANICAL  DEFINITIONS 


35 


65.  A  carrier,  intermediate,  or  idle,  gear  is  a  gear 
running  on  a  stud  used  either  to  change  the  direction  of 
rotation  of  a  driven  gear  or  to  connect  a  driving  and  driven 
shaft  when  gears  of  sufficiently  large  diameter  cannot  be 
used  to  make  the  connection  directly.  When  one  inter¬ 
mediate  is  used,  the  direction  of  rotation  of  the  driven  gear 
is  the  same  as  that  of  the  driver;  but  when  two  intermediates 
are  used,  the  direction  of  rotation  is  the  reverse,  as  though 
the  driver  meshed  directly  with  the  driven  gear. 


SCREWS 

66.  A  screw  is  a  straight  shaft  with  a  thread  cut 
spirally  around  it.  It  may  have  a  right-hand  or  a  left-hand 
thread,  depending  on  whether  it  advances  when  turned  to 
the  right  or  to  the  left. 

The  pitch  of  a  screw  is  expressed  in  threads  per  inch  for 
small  screws,  but  for  large  screws  it  is  usually  expressed  as 
the  distance  between  the  centers  of  consecutive  turns  of  the 
thread  measured  parallel  to  the  axis  of  the  screw. 

67.  A  multiple-threaded  screw  is  one  with  two  or 
more  threads  cut  on  it. 

68.  A  reciprocating  screw  is  one  with  both  a  right- 
hand  and  a  left-hand  thread  cut  on  it,  the  threads  being  con¬ 
nected  at  each  end  and  designed  to  engage  with  a  pin,  or 
some  form  of  traveler,  for  the  purpose  of  imparting  a 
reciprocating  motion. 

69.  A  differential  screw  is  one  with  a  varying  pitch 
and  is  designed  to  engage  with  a  pin,  as  it  is  obviously 
impossible  to  fit  such  a  screw  with  a  threaded  nut. 


CAMS 

70.  A  cam  is  a  turning  or  sliding  piece  that,  by  the  shape 
of  its  curved  edge  or  face,  or  a  groove  in  its  surface,  imparts  a 
variable  or  intermittent  motion  to  a  roller,  lever,  rod,  or  other 
moving  part,  the  character  of  the  imparted  motion  depending 
on  the  shape  of  the  cam.  This  motion,  where  a  small  amount 
of  power  is  involved,  is  often  transmitted  by  sliding  contact; 


MECHANICAL  DEFINITIONS 


36 


§7 


but  in  cases  where  much  force  is  employed,  the  contact  is 
generally  rolling. 

The  lieel  of  a  cam  is  that  part  of  its- effective  circumfer¬ 
ence  that  is  nearest  the  axis  of  rotation.  The  toe  of  a  cam 
is  the  part  of  its  effective  circumference  that  is  farthest  from 
the  axis  of  rotation.  The  difference  in  the  distance  between 
the  toe  of  the  cam  from  the  axis  of  rotation  and  the  heel  of 

the  cam  from  the  axis  of  rotation  is 
called  the  throw  of  the  cam. 

71.  A  positive-motion  cam  is 
one  that  actuates  the  part  that  is  in 
direct  contact  with  it  equally  well  in  all 
directions  without  the  aid  of  a  spring 
or  weight;  a  common  type  is  shown 
in  Fig.  39.  It  consists  of  a  plate  a  fast¬ 
ened  to  a  rotating  shaft  and  having  a 
groove  b  cut  in  one  face.  A  roller  c 
attached  to  a  three-armed  lever  d 


engages  with  the  groove;  thus,  as  the  cam  rotates,  it  imparts 
to  the  lever  a  motion  that  depends  on  the  shape  of  the  guid¬ 
ing  groove.  The  lever  d  is  moved  positively  in  both  direc¬ 
tions  and  nothing  less  than  actual  breakage  of  some  part  can 
prevent  its  working. 

72.  A  non-positive  cam  is  one  that  depends  on  either 
a  spring  or  a  weight  to  keep  the  part  on  which  it  acts  in 
direct  contact  with  it  at  all  times;  a  non-positive  cam  is 


§7 


MECHANICAL  DEFINITIONS 


37 


shown  in  Fig.  40.  Like  the  positive-motion  cam,  it  consists 
of  a  plate  a  fastened  on  a  rotating  shaft,  but  the  roller  c, 
instead  of  being  guided  by  a  groove,  rests  on  the  circum¬ 
ference  of  the  cam.  Motion  will  be  positively  imparted  to 
the  rod  e  only  while  the  cam-bowl,  or  roller,  is  traversing 
from  the  heel  to  the  toe  of  the  cam.  While  passing  from  the 
toe  to  the  heel,  the  roller  is  held  in  contact  with  the  cam  by 
the  spring  b  attached  to  the  lever  d.  If  the  rod  e  or  lever  d 
is  caught  on  the  return  motion  so  that  the  tension  of  the 
spring  is  overcome,  the  roller  c  will  be  held  away  from  the 
cam. 

73.  A  cam-bowl  is  a  small  roller  that  runs  either  in  the 
groove  of  a  positive  cam  or  on  the  circumference  of  a  non¬ 
positive  cam,  in  the  latter  case  being  held  in  position  by  a 
spring  or  weight. 

74.  If  the  action  of  a  cam  is  intermittent,  it  is  called 

a  wiper.  Occasionally  the  term  wiper  is  applied  to  any 
non-positive  cam.  _ 


LEVERS 

75.  A  lever  is  an  inflexible  bar  capable  of  being  freely 
moved  about  a  fixed  point,  or  line,  called  the  fulcrum. 
The  bar  is  acted  on  at  two  points  by  two  forces  that  tend  to 
rotate  it  in  opposite  directions  about  its  fulcrum.  Of  these 
two  forces,  the  one  that  is  applied  with  the  purpose  of 


Fig.  41 


imparting  motion  is  termed  the  power ,  while  the  force  that 
is  to  be  overcome  is  the  weight ,  or  resistance.  The  parts  a  f 
and  b  /,  Fig.  41,  are  the  arms  of  the  lever. 

There  are  three  classes,  or  varieties,  of  levers  depending 
on  the  relative  positions  of  the  power  p ,  weight  w,  and  ful¬ 
crum  /. 


38 


MECHANICAL  DEFINITIONS 


§7 


If  the  fulcrum  is  between  the  power  and  the  weight  ( p ,  /,  w) , 
as  shown  in  Fig.  41,  the  lever  is  of  the  first  class.  In  this 
combination  equilibrium  exists  if  the  product  of  the  force  p 
times  arm  a  f  equals  the  product  of  the  force  w  times  arm  b  /. 

l-/  _ 

/«} 

Fig.  42 

If  the  weight  is  between  the  power  and  the  fulcrum  ( p ,  w,  /) 
as  shown  in  Fig.  42,  the  lever  is  of  the  second  class. 

If  the  power  is  between  the  weight  and  the  fulcrum  ( w,p,  /) , 
the  lever  is  of  the  third  class,  Fig.  43. 


Fig. 43 


76.  Sometimes  it  is  not  convenient  to  use  a  lever  suffi¬ 
ciently  long  to  make  a  given  power  support  a  given  weight. 
In  this  case,  combinations  of  levers  known  as  compound 
level’s  are  used. 


MECHANICAL  CALCU¬ 
LATIONS 


RULES  PERTAINING  TO  THE  TRANS¬ 
MISSION  OF  POWER 


RULES  APPLYING  TO  SHAFTING 

1.  Cold-Rolled  Shafting. — The  following  rules  will  be 
found  useful  in  finding  the  required  size  of  a  cold-rolled 
shaft  necessary  to  transmit  a  given  horsepower. 

To  find  the  required  diameter  of  a  main  shaft: 

Rule  I. — Find  the  cube  root  of  100  times  the  required  horse¬ 
power  divided  by  the  desired  number  of  revolutions  of  the  shaft 
per  minute. 

To  find  the  required  diameter  of  line  shafts  to  transmit  a 
given  horsepower  with  the  power  taken  off  at  intervals  and 
the  bearings  of  the  shaft  not  more  than  8  feet  apart: 

Rule  II. — Find  the  cube  root  of  50  times  the  required  horsc- 
pozver  divided  by  the  desired  number  of  revolutions  per  minute. 

To  find  the  required  diameter  of  short  countershafts  for 
transmitting  a  given  horsepower: 

Rule  III. — Find  the  cube  root  of  30  times  the  required 
horsepower  divided  by  the  desired  number  of  revolutions  per 
minute. 


For  notice  of  copyright,  see  page  immediately  following  the  title  page 


2 


MECHANICAL  CALCULATIONS 


§8 


Example. — Suppose  that  it  is  desired  to  purchase  a  line  shaft  for  a 
weave  room  requiring  350  horsepower;  it  is  desired  to  have  the  shaft 
make  300  revolutions  per  minute  and  a  cold-rolled  shaft  is  to  be  used. 
What  diameter  of  shafting  is  required? 

50  X  H  P 

Solution. — Diameter  of  shaft  equals  cube  root  of 


rev.  oer  mm. 


50X  350 
300 


=  3.87+  in.  Ans. 


Note.— In  a  case  like  this  a  4-inch  cold-rolled  shaft  would  probably  be  ordered'  as 
this  would  allow  for  the  extra  power  required  to  overcome  the  friction  of  the  shaft  in 
its  bearings. 


2.  Turned  Shafting. — The  following  rules  give  the 
methods  of  finding  the  required  size  of  turned  shafting  to 
transmit  a  required  horsepower. 

To  find  the  required  diameter  of  a  main  shaft: 


Rule  I. — Find  the  cube  root  of  125  times  the  required  horse¬ 
power  divided  by  the  desired  number  of  revolutions  per  minute. 

To  find  the  required  diameter  of  line  shafts  with  the 
power  taken  off  at  intervals  and  the  bearings  not  more  than 
8  feet  apart: 


Rule  II. — Find  the  cube  root  of  90  times  the  reqziired  horse- 
pozver  divided  by  the  desired  number  of  revolutions  per  minute. 

To  find  the  required  diameter  of  short  countershafts: 


Rule  III. — Find  the  cube  root  of  50  times  the  required  horse¬ 
power  divided  by  the  desired  number  of  revolutions  per  minute. 

Example.— What  diameter  of  turned  shafting  is  capable  of  trans¬ 
mitting  45  horsepower,  the  shaft  to  be  the  main  driving,  or  line,  shaft 
of  the  room  and  the  bearings  not  more  than  8  feet  apart.  It  is  desired 
that  the  shaft  make  150  revolutions  per  minute. 

90  X  H  P 

Solution. — Diameter  of  shaft  equals  cube  root  of - : — —• 

rev.  per  mm. 

J90  X  45  =  3  in  Ans 
\  150 

Note.— When  the  hangers  are  placed  far  apart,  a  larger  shaft  is  necessary  in 
order  that  it  may  have  stiffness  to  withstand  the  bending  strain  due  to  its  lack  of  sup¬ 
port  and  to  its  own  weight. 


8 


MECHANICAL  CALCULATIONS 


3 


RULES  APPLYING  TO  SPEEDS 

3.  Speeds  of  Pulleys. — In  connection  with  pulleys  it 
must  be  remembered  that  the  driving  pulley  is  the  one  that 
furnishes  the  power  to  the  driven  pulley.  The  tight  side  of 
the  belt  always  travels  toward  the  driving  pulley  and  the 
slack  side  toward  the  driven  pulley. 

To  find  the  number  of  revolutions  per  minute  of  the 
driven  pulley,  the  diameter  and  number  of  revolutions  of 
the  driving  pulley  and  the  diameter  of  the  driven  pulley 
being  known: 

Rule  I. — Multiply  the  diameter  of  the  driving  pulley  by  its 
revolutions  and  divide  the  product  by  the  diameter  of  the  driven 
pulley . 

Example  1. — A  driving  shaft  making  350  revolutions  per  minute 
carries  a  24-inch  pulley  that  drives  a  14-inch  pulley  on  the  main  shaft 
of  a  machine;  find  the  revolutions  of  the  main  shaft  of  the  machine. 

24  in.  X  3.50  . 

Solution. —  - — — - =  600  rev.  per  mm.  Ans. 

14  m.  v 

To  find  the  revolutions  of  the  driving  shaft,  the  diameter 
and  revolutions  of  the  driven  pulley  and  the  size  of  the  dri¬ 
ving  pulley  being  known: 

Rule  II. — Multiply  the  diameter  of  the  driven  pulley  by  its 
speed  and  divide  by  the  diameter  of  the  driving  pulley. 

Example  2. — The  shaft  of  a  machine  makes  700  revolutions  per 
minute.  The  size  of  the  driven  pulley  is  8  inches  and  the  driving  pul¬ 
ley  on  the  main  shaft  is  14  inches;  find  the  revolutions  of  the  main 
driving  shaft. 

8  in.  X  700 

Solution. —  — 7—; - =  400  rev.  per  mm.  Ans. 

14  in.  ^ 

To  find  the  diameter  of  a  driven  pulley  to  give  any  desired 
number  of  revolutions  per  minute,  the  diameter  of  the  driving 
pulley  and  its  speed  being  known: 

Rule  III. — Multiply  the  diameter  of  the  driving  pulley  by 
its  speed  and  divide  the  product  by  the  desired  number  of  revolu¬ 
tions  of  the  driven  pulley. 

Example  3. — The  main  shaft  of  a  room  makes  225  revolutions  per 
minute  and  carries  a  20-inch  pulley  from  which  it  is  desired  to  drive 


4 


MECHANICAL  CALCULATIONS 


§8 


a  countershaft  300  revolutions  per  minute;  what  size  pulley  must  be 
ordered  for  the  countershaft? 

20  v  225 

Solution. —  .  —  =  15-in.  pulley.  Ans. 

oUU 

To  find  the  diameter  of  a  driving  pulley  necessary  on  a 
shaft  making  a  definite  number  of  revolutions  in  order  to 
drive  a  driven  pulley  of  a  given  diameter  a  certain  speed; 

Rule  IV. — Multiply  the  diameter  of  the  driven  pulley  by  the 
desired  speed  and  divide  the  product  by  the  speed  of  the  driving 
shaft. 

Example  4. — Find  the  size  of  the  pulley  required  on  a  driving  shaft 
making  360  revolutions  per  minute  in  order  to  drive  a  machine  600  revo¬ 
lutions  per  minute.  The  size 
of  the  driven  pulley  on  the 
machine  is  12  inches. 
Solution. — 

12  X  600  „ 

360 —  =  20-in.  pulley.  Ans. 

4.  Effect  of  Coun¬ 
ter  s  li  a  f  t  s  on  Speed. 

It  often  happens  that 
the  power  is  transmitted 
through  a  countershaft, 
or  countershafts,  carrying 
different-sized  pulleys  be¬ 
fore  being  applied  to  the 
pulley  whose  speed  it  is 
desired  to  find. 

To  find  the  speed  of  a 
driven  pulley  when  the 
power  is  transmitted  through  one  or  more  countershafts 
carrying  different-sized  pulleys: 

Rule. — Multiply  the  speed  of  the  driving  shaft  by  the  product 
of  the  diameters  of  all  the  driving  pulleys  and  divide  the  result 
by  the  product  of  the  diameters  of  all  the  driven  pulleys. 

Example. — Referring  to  Fig.  1,  suppose  that  the  driving  shaft 
makes  375  revolutions  per  minute  and  that  the  main  driving  pulley  a 
is  18  inches  in  diameter  and  drives  a  12-inch  pulley  £  on  a  countershaft. 


MECHANICAL  CALCULATIONS 


5 


On  this  countershaft  a  22-inch  pulley  c  drives  the  10-inch  pulley  d  of 
a  machine.  Find  the  number  of  revolutions  of  the  pulley  d. 


Solution. — 


375  X  18  in.  X  22  in. 
12  in.  X  10  in. 


1,237.5  rev.  per  min. 


Ans. 


5.  Ratios  of  the  Speeds  and  the  Diameters  of 
Pulleys. — From  these  rules  it  will  be  seen  that  speeds  and 
variations  of  speeds  and  the  diameters  of  driving  and  driven 
pulleys  should  be  treated  as  ratios  and  that  the  following 
equality  of  ratios,  or  proportion,  is  always  true;  namely,  that 
the  diameter  of  the  driving  pulley  is  to  the  diameter  of  the 
driven  pulley  as  the  speed  of  the  driven  pulley  is  to  the  speed 
of  the  driving  pulley.  As  the  product  of  the  extremes  is 
always  equal  to  the  product  of  the  means  in  a  proportion,  so 
the  diameter  of  the  driving  pulley  multiplied  by  its  speed 
always  equals  the  diameter  of  the  driven  pulley  multiplied 
by  its  speed. 


6.  To  find  the  surface  velocity  of  a  rotating  pulley  or 
cylinder  or  the  speed  of  a  belt  passing  around  it,  in  feet 
per  minute  (slip  neglected): 

Rule. — I.  Multiply  the  diameter  of  the  pulley  or  cylinder 
in  feet  by  3.1416  and  by  the  number  of  revolutions  per  minute. 

II.  If  the  diameter  of  the  pulley  or  other  cylinder  is  expressed 
in  inches ,  multiply  its  diameter  by  3.1416  and  by  the  member  of 
revolutions  per  minute  that  it  makes  and  divide  the  product  by  12. 

Example.— -Find  the  surface  velocity,  in  feet  per  minute,  of  a 
50-inch  cylinder  making  160  revolutions  per  minute. 

c  50  X  3.1416  X  160  onni4Ii  .  . 

Solution. —  - ^ -  =  2,094.4  ft.  per  mm.  Ans. 

J -Pj 


EXAMPLES  FOR  PRACTICE 

1.  Required  the  speed  of  a  driven  shaft  on  which  there  is  a  14-inch 

pulley,  the  power  being  transmitted  by  a  22-inch  pulley  making 
280  revolutions  per  minute.  Ans.  440  rev.  per  min. 

2.  What  size  pulley  should  be  placed  on  a  driving  shaft  making 
320  revolutions  per  minute  in  order  to  drive  a  machine  500  revolutions 
per  minute,  the  size  of  the  driven  pulley  being  12  inches? 

Ans.  18f-in.  pulley 


6 


MECHANICAL  CALCULATIONS 


§8 


3.  What  size  pulley  is  required  on  a  countershaft  that  is  to  be 

driven  160  revolutions  per  minute  from  a  24-inch  pulley  on  a  main 
shaft  making  220  revolutions  per  minute?  Ans.  33-in.  pulley 

4.  A  main  shaft  carries  a  28-inch  pulley  and  drives  a  21-inch  pulley 
700  revolutions  per  minute;  what  is  the  speed  of  the  main  shaft? 

Ans.  525  rev.  per  min. 

5.  A  20-foot  flywheel  makes  60  revolutions  per  minute  and  drives 
a  6-foot  pulley  on  the  head-shaft.  A  5-foot  pulley  on  the  head-shaft 
drives  a  4-foot  pulley  on  the  main  shaft  of  a  weave  room.  On  the 
main  shaft  of  the  weave  room  a  2-foot  pulley  drives  a  2^-foot  pulley 
on  a  countershaft.  Find  the  speed  of  the  countershaft. 

Ans.  200  rev.  per  min. 

Note.— In  these  Examples  for  Practice  it  will  be  found  an  advantage  in  order  to 
better  understand  any  example  to  make  a  rough  sketch  of  the  arrangement  of  pulleys 
and  then  designate  the  driving  and  driven  pulleys. 


RULES  APPLYING  TO  BELTS 

7.  Length  of  Belts. — The  following  rules  will  enable 
the  student  to  perform  calculations  in  connection  with  belts. 

To  find  the  length  of  an  open  belt,  the  distance  between 
the  centers  of  the  shafts  and  the  diameters  of  the  driving  and 
driven  pulleys  being  known: 

Rule  I. — Multiply  half  the  sum  of  the  diameters  of  the 
driving  and  driven  pulleys  by  3.1416  and  to  this  product  add 
twice  the  distance  between  centers. 

Example  1.— A  countershaft  is  to  be  driven  from  the  main  shaft 
with  an  open  belt;  the  distance  between  the  centers  of  the  shafts  is 
12  feet,  and  the  diameters  of  the  driving  and  driven  pulleys  are, 
respectively,  2  and  3  feet;  how  long  a  belt  is  required? 

Solution.—  ~1  =  2.5;  2.5  X  3.1416  =  7.854 

12  X  2  =  24;  24  +  7.854  =  31.854  ft.  of  belt.  Ans. 

Note— In  case  one  pulley  is  much  larger  than  the  other,  it  is  well  to  cut  the  belt 
2  or  3  inches  longer  than  calculated  by  the  above  rule. 

To  find  the  length  of  a  crossed  belt,  the  size  of  the  driving 
and  driven  pulleys  and  the  distance  between  the  centers  of 
the  shafts  being  known: 

Rule  II. — To  one-half  the  Product  of  the  sum  of  the  diameters 
of  the  driving  and  driven  pulleys  and  3.1416  add  twice  the 


§8 


MECHANICAL  CALCULATIONS 


7 


square  root  of  the  sum  of  the  square  of  the  distance  between  the 
centers  of  the  shafts  and  the  square  of  one-half  the  sum  of 
the  diameters  of  the  driving  and  driven  pulleys. 

Example  2. — A  countershaft  is  to  be  driven  from  the  main  shaft 
with  a  crossed  belt,  the  distance  between  the  centers  of  the  shafts  is 
12  feet,  and  the  diameters  of  the  driving  and  driven  pulleys  are, 
respectively,  2  and  3  feet;  how  long  a  belt  is  required? 

Solution. — 


(2  +  3)  X  3.1416 
2 


2  X  Vl44  +X+  = 

2  X  Vl44  +  6.25  = 

2  X  VloO.25  = 

2  X  12.25  =  24.5 

24.5  +  7.854  =  32.354  ft.  of  belt.  Ans. 


Note.— These  rules,  although  not  absolutely  accurate,  are  near  enough  for  prac¬ 
tical  purposes  when  it  is  impossible  to  measure  the  length  of  belt  required. 


8.  Horsepower  Transmitted  by  Belts. — As  the  width 
of  a  belt  required  to  transmit  a  given  horsepower  depends 
on  the  speed  and  tension  of  the  belt,  the  size  of  the  smaller 
pulley,  and  the  relative  amount  of  its  surface  touched  by  the 
belt,  no  rule  can  be  given  that  will  apply  to  all  cases. 
A  belt  that  is  being  constantly  shifted  from  a  tight  to  a 
loose  pulley,  or  vice  versa,  must  be  wider  than  one  running 
on  the  same  pulley  all  the  time,  while  innumerable  other 
conditions  govern  the  horsepower  capable  of  being  trans¬ 
mitted  by  a  given  belt  and  the  life  of  the  belt. 

It  has  been  found  by  exhaustive  experiments  that  a  single 
belt  traveling  900  feet  per  minute  will  transmit  approxi¬ 
mately  1  horsepower  per  inch  of  width  when  the  arc  of  con¬ 
tact  on  the  smaller  pulley  does  not  vary  much  from  180°. 
This  is  used  by  many  engineers  as  a  general  law  for  belting 
and  is  applied  in  all  cases.  From  this  fact  the  following 
rules  in  connection  with  belting  are  obtained. 

To  find  the  horsepower  transmitted  by  a  given  belt: 

Kule  I. — Divide  the  product  of  the  width  of  the  belt ,  in 
inches ,  and  the  speed ,  in  feet  per  minute,  by  900. 


8 


MECHANICAL  CALCULATIONS 


8 


To  find  the  required  width  of  a  belt  to  transmit  a  given 
horsepower: 

llule  II. — Divide  the  horsepower  multiplied  by  900  by  the 
speed  of  the  belt ,  in  feet  per  minute. 

Example. — Two  48-inch  pulleys  are  to  be  connected  by  a  single 
belt  and  make  200  revolutions  per  minute;  if  40  horsepower  is  to  be 
transmitted  what  must  be  the  width  of  the  belt? 


200  X  48  X  3.1416 
12 


=  2,513  ft.  per 


Solution. —  The  belt  speed  =  — 


Ans. 


Note.— A  14-inch  belt  might  safely  be  used,  since  the  rule  gives  a  liberal  width 
when  the  pulleys  are  of  equal  size. 

9.  In  these  rules  it  has  been  assumed  that  the  belt  is 
open  and  also  that  the  driving  and  driven  pulleys  are  of  the 
same  diameter,  the  belt  consequently  being  in  contact  with 
half  of  the  circumference  of  each  pulley.  But  when  one 
pulley  is  larger  than  the  other,  the  horsepower  transmitted 
is  reduced  as  the  arc  of  the  smaller  pulley  that  is  in  contact 
with  the  belt  is  reduced.  With  a  crossed  belt  the  amount  of 
horsepower  that  can  be  transmitted  by  a  given  width  of  belt 
is  increased,  as  there  is  then  more  of  the  surface  of  the 
pulleys  in  contact  with  the  belt. 

As  the  rules  for  single  belts  are  based  on  the  strength 

at  the  lace  holes,  a  double  belt,  which  is  twice  as  thick, 

/■ 

should  be  able  to  transmit  twice  as  much  power  as  a  single 
belt  and,  in  fact,  more  than  this  where,  as  is  quite  common, 
the  ends  of  the  belt  are  cemented  together  instead  of  being 
laced.  Where  double  belts  are  used  on  small  pulleys,  how¬ 
ever,  the  contact  with  the  pulley  face  is  less  nearly  perfect  than 
it  would  be  if  a  single  belt  were  used,  owing  to  the  greater 
rigidity  of  the  former.  More  work  is  also  required  to  bend 
the  belt  as  it  runs  over  the  pulley  than  in  the  case  of  the 
thinner  and  more  pliable  belt,  and  the  centrifugal  force 
tending  to  throw  the  belt  from  the  pulley  also  increases  with 
the  thickness.  For  these  reasons,  the  width  of  a  double 
belt  required  to  transmit  a  given  horsepower  is  generally 
assumed  to  be  seven-tenths  the  width  of  a  single  belt 


§8 


MECHANICAL  CALCULATIONS 


9 


required  to  transmit  the  same  power.  Therefore,  in  order 
to  find  the  width  of  a  double  belt  required  to  transmit  a 
given  horsepower,  proceed  as  with  a  single  belt  and  multiply 
the  result  by  tV;  and  in  order  to  find  the  horsepower  trans¬ 
mitted  by  a  given  width  of  a  double  belt,  proceed  as  with  a 
single  belt  and  multiply  the  result  by  -7°. 


ROPE  TRANSMISSION 

10.  Rope  drives  give  the  most  satisfactory  results  when 
the  ropes  run  in  grooves  the  sides  of  which  have  an  angle 
of  about  45°. 

To  find  the  horsepower  transmitted  by  a  single  rope  run¬ 
ning  under  favorable  conditions  in  a  45°  groove: 


Rule. — Multiply  the  speed  of  the  rope,  in  feet  per  second ,  by 
the  square  of  its  diameter ,  in  inches ,  and  divide  the  product 
by  825.  This  quotient  multiplied  by  the  result  obtained  by  sub¬ 
tracting  from  200  the  speed  of  the  rope  per  second ,  squared ,  and 
divided  by  107.2  equals  the  horsepower  that  can  be  transmitted 
with  a  single  rope. 

Example. — A  flywheel  designed  for  a  rope  drive  is  22  feet  in  diam¬ 
eter  and  is  equipped  with  30  grooves;  the  diameter  of  the  rope  is 
li  inches  and  the  flywheel  makes  50  revolutions  per  minute;  what 
horsepower  can  be  safely  transmitted? 

Solution.— 


22  X  3.1416  X  50 
60  (sec.) 

57.596  X  (li) 


825 

57.596  X  1.5625 


825 


=  57.596,  speed  of  rope  in  ft.  per  sec. 
3317.299216^  _ 


X 


^200  - 


107.2 

.109083  X  (200  -  30.9449) 


.109083  X  169.0551  =  18.441,  H.  P.  for  one  rope.  Since  there  are 
30  grooves  in  the  flywheel,  in  each  of  which  there  is  one  rope,  the 
total  power  transmitted  will  be  18.441  X  30  =  553.23  H.  P.  Ans. 


Fig.  2  shows  the  horsepower  transmitted  by  1-inch,  li-inch, 
li-inoh,  lf-inch,  and  2-inch  ropes  for  various  velocities.  The 
horizontal  distances  represent  velocities  in  feet  per  second, 


10 


MECHANICAL  CALCULATIONS 


§8 


and  the  vertical  distances  the  horsepower  transmitted  by  a 
single  rope.  It  shows  that  the  maximum  power  is  obtained 
at  a  speed  of  about  84  feet  per  second.  For  higher  veloci¬ 
ties,  the  centrifugal  force  becomes  so  great  that  the  power  is 
decreased,  and  when  the  speed  reaches  145  feet  per  second. 


Pig.  2 


the  centrifugal  force  just  balances  the  tension,  so  that  no 
power  at  all  is  transmitted.  Consequently,  a  rope  should  not 
run  faster  than  about  5,000  feet  per  minute,  while  it  is  prefer¬ 
able,  on  the  score  of  durability,  to  limit  the  velocity  to  3,500 
feet  per  minute. 


RULES  APPLYING  TO  GEARS 

11.  The  rules  that  apply  to  pulleys  are  also  applicable  to 
gears  with  the  exception  that  the  number  of  teeth  on  the 
gears  are  taken  instead  of  the  diameters. 

To  find  the  speed  of  a  driven  gear: 

Rule. — Multiply  the  speed  of  the  first  driving  gear  by  its 
number  of  teeth  and  by  the  number  of  teeth  on  each  driving  gear 
in  the  train,  if  there  is  more  than  one ,  and  divide  the  Product  by 


§8 


MECHANICAL  CALCULATIONS 


II 


the  number  of  teeth  on  the  driven  gear  or  by  the  product  of  the 
teeth  on  the  driven  gears. 

Example. — Suppose  that  the  shaft  e  in  Fig.  3  is  the  driving  shaft 
and  makes  40  revolutions  per  minute;  find  the  number  of  revolutions 
of  the  driven  shaft  f  when  a  and  c  have  each  24  teeth  and  d  and  b 
11  teeth. 

c  40  X  24  X  24 

Solution. —  — 11  1  — -  =  190.413  rev.  per  min.  Ans. 

11  y\  11 

A  good  method  of  determining  whether  a  gear  is  a  driver 
or  driven  gear  is  to  notice  which  side  of  its  teeth  are  worn, 
or  polished  smooth.  The  driving  gear  always  has  its  teeth 
polished  on  the  side  facing  in  the  direction  in  which  the  gear 
is  moving;  a  driven  gear  has  the  opposite  side  of  the  teeth 


polished;  while  on  an  intermediate  gear  both  sides  of  the 
teeth  are  worn. 

For  solving  problems  that  deal  with  gears,  use  the  same 
rules  as  are  given  for  pulleys,  remembering  that  the  number 
of  teeth  on  the  driving  gear  or  gears  multiplied  together  and 
by  the  speed  of  the  first  driver  equals  the  number  of  teeth  on 
the  driven  gear  or  gears  times  the  speed  of  the  last  driven 
gear.  Speeds  and  sizes  of  gears,  like  pulleys,  should  be 
treated  by  proportion.  Intermediate  gears  should  not  be 
used  when  finding  speeds  or  sizes  of  gears. 

Where  two  gears  are  fastened  together  and  interposed 
between  driving  and  driven  gears  after  the  manner  of  an 
intermediate,  they  are  said  to  be  compounded  or  to  constitute 


12 


MECHANICAL  CALCULATIONS 


§8 


a  compound.  If  the  two  gears  vary  in  size,  such  an  arrange¬ 
ment  will  affect  the  value  of  the  train  and  they  should  be 
figured  as  a  driver  and  a  driven  gear,  as  they  really  are. 

12.  Sizing:  Gear-Blanks. — Many  mills  are  equipped 
with  gear-cutting  machines  for  cutting  gears  to  replace 
broken  or  worn  ones,  making  change  gears,  etc.  When  it  is 
desired  to  cut  a  gear,  it  is  necessary  to  select  a  gear-blank  of 
the  correct  diameter  for  the  desired  number  of  teeth  and  pitch. 

To  find  the  desired  diameter  of  blank  for  any  number  of 
teeth  and  any  diametral  pitch: 

Rule  I. — Add  2  to  the  required  number  of  teeth  and  divide  by 
the  desired  pitch;  the  quotient  is  the  diameter ,  expressed  in 
inches ,  of  the  blank  required. 

Example  1.— A  change  gear  with  33  teeth  and  10-pitch  is  desired. 
To  what  diameter  must  the  gear-blank  be  turned? 

33  +  2 

Solution. —  ~To”  =  ^  *n-  ^ns- 

To  find  the  number  of  teeth  the  gear-cutter  must  space  to 
cut  a  given  blank  a  required  pitch: 

Rule  II. — Multiply  the  diameter  of  the  blank  by  the  required 
pitch  and  from  the  product  thus  obtained  subtract  2;  the  answer 
is  the  number  of  teeth  required. 

Example  2. — How  many  teeth  must  the  gear-cutter  space  to  cut 
a  gear-blank  21  inches  in  diameter,  12-pitch? 

Solution. —  2\  X  12  =  30;  30  —  2  =  28  teeth.  Aus. 

13.  Speed  of  Worm-Gears. — To  find  the  speed  of  a 
worm-gear  driven  by  a  single-  or  double-threaded  worm: 

Rule  I. — If  the  worm  is  single-threaded  divide  its  speed  by 
the  number  of  teeth  in  the  worm-gear. 

II.  If  the  worm  is  double-threaded  multiply  its  speed  by  2 
and  divide  the  product  by  the  number  of  teeth  in  the  worm-gear. 

Example  1. — A  00-tooth  worm-gear  is  driven  by  a  single-threaded 
worm  making  300  revolutions  per  minute;  find  its  number  of  revolu¬ 
tions  per  minute. 


Solution. — 


— -  =  5  rev.  per  min.  Ans. 
60 


8 


MECHANICAL  CALCULATIONS 


13 


Example  2. — An  80-tooth  worm-gear  is  driven  by  a  double-threaded 
worm  making  160  revolutions  per  minute;  find  the  number  of  revolu¬ 
tions  per  minute  of  the  worm-gear. 

c  160  X  2  A 

Solution. —  — -- —  =  4  rev.  per  mm.  Ans. 

oU 

A  worm  is  always  a  driver  and  reckoned  as  a  one-tooth 
gear  if  single-threaded  and  a  two-tooth  gear  if  double- 
threaded. 

14.  A  mangle  gear  is  always  a  driven  gear  and  its  speed 
is  found  in  the  same  manner  as  that  of  an  ordinary  gear 
except  that  its  size  is  reckoned  as  somewhat  larger  than 
it  really  is,  because  the  driving  pinion,  while  rounding  each 
end  of  the  row  of  pegs,  makes  a  half-revolution,  which 
moves  the  mangle  but  one  peg. 

To  find  how  many  complete  cycles,  or  double  revolutions, 
a  mangle  gear  will  make  per  minute: 

Rule. — Divide  the  product  of  the  number  of  revolutions  per 
minute  of  the  driving  pinion  and  the  number  of  teeth  that  it  con¬ 
tains  by  the  sum  of  twice  the  number  of  pegs  in  the  mangle  and 
the  number  of  teeth  in  the  driving  pinion  diminished  by  2. 

Example. — A  10-tooth  pinion  gear  making  216  revolutions  per 
minute  drives  a  mangle  gear  with  176  teeth,  or  pegs;  how  many  com¬ 
plete  cycles  per  minute  does  the  mangle  gear  make? 

0  216  X  10  216  X  10  216  X  10 

{solution.  (176  X  2)  +  (10  -  2)  ~  352  +  8  ~  360 

complete  cycles.  Ans. 

That  is,  the  mangle  gear  would  make  12  revolutions,  6  in  one  direc¬ 
tion  and  6  in  the  other. 


EXAMPLES  FOR  PRACTICE 

1.  A  shaft  making  30  revolutions  per  minute  must  drive  a  shaft 

carrying  a  40-tooth  gear  21  revolutions  per  minute;  what  gear  is 
required  as  a  driver?  Ans.  28-tooth  gear 

2.  A  driving  gear  with  60  teeth  makes  70  revolutions  per  minute 
and  drives  a  pinion  gear  of  18  teeth;  find  the  speed  of  the  pinion  gear. 

Ans.  233^  rev.  per  min. 

3.  The  speed  of  the  first  driving  shaft  is  40  revolutions  per  minute; 

the  driving  gears  contain  48,  60,  and  40  teeth,  respectively;  the  driven 
gears,  40,  48,  and  40  teeth,  respectively;  find  the  speed  of  the  last 
driven  shaft.  Ans.  60  rev.  per  min. 


14 


MECHANICAL  CALCULATIONS 


§8 


4.  A  double-threaded  worm  making  300  revolutions  per  minute 

drives  an  80-tooth  worm-gear;  how  many  revolutions  per  minute  does 
the  worm-gear  make?  Ans.  7i  rev.  per  min. 

5.  A  mangle  gear  having  116  teeth  is  driven  by  a  10-tooth  pinion 
gear  making  120  revolutions  per  minute;  how  many  double  revolutions 
per  minute  does  the  mangle  gear  make?  Ans.  5  double  revolutions 

6.  A  train  of  gears  is  composed  of  four  drivers  containing  32,  24, 
40,  and  16  teeth,  respectively,  and  four  driven  gears  containing  24,  16, 
32,  and  8  teeth,  respectively;  the  first  driver  makes  16  revolutions  per 
minute;  find  the  speed  of  the  last  driven  shaft.  Ans.  80  rev.  per  min. 

7.  A  72-tooth  worm-gear  is  driven  by  a  double-threaded  worm 

making  198  revolutions  per  minute;  find  the  number  of  revolutions  per 
minute  of  the  worm-gear.  Ans.  5i  rev.  per  min. 

8.  A  6-tooth  pinion  making  265  revolutions  per  minute  drives  an 

83-tooth  mangle  gear;  how  many  complete  cycles  does  the  mangle 
gear  make?  Ans.  9  cycles 


CONSTANTS 

15.  It  often  happens  that  though  a  machine  contains  a 
more  or  less  complicated  train  of  gears,  only  one  of  them  is 
changed  for  alterations  in  the  speed  of  the  driven  gear.  This 
gear  is  known  as  a  change  gear,  and  the  arrangement  of 
the  train  is  such  that  its  size  may  readily  be  changed  without 
disturbing  the  other  members  of  the  train.  If  the  change 
gear  is  a  driven  gear,  an  increase  in  its  size  will  diminish 
the  speed  of  the  driven  gear  or  shaft.  If  it  is  a  driver,  an 
increase  in  its  size  will  increase  the  speed  of  the  driven  gear 
or  shaft. 

Where  a  train  of  gears  is  employed,  the  calculation  of  the 
required  size  of  change  gear  to  produce  a  given  speed  of  the 
driven  gear  becomes  rather  long  unless  some  method  of 
shortening  the  operation  is  adopted.  This  may  be  accom¬ 
plished  by  partly  performing  the  operation  and  securing  a 
number  that  expresses  the  value  of  the  train,  excluding  the 
change  gear,  and  that  needs  but  one  multiplication  or  divi¬ 
sion  to  obtain  the  desired  speed  or  the  desired  size  of  change 
gear;  such  a  number  is  called  a  constant. 

16.  A  constant  number  may  be  either  a  constant  factor 
or  a  co?istant  dividend.  A  constant  factor  is  a  number 


§8 


MECHANICAL  CALCULATIONS 


15 


that,  when  multiplied  by  the  change  gear,  gives  the  speed 
of  the  driven  shaft  of  a  train  of  gears  and,  when  divided  into 
the  speed  of  the  driven  shaft,  gives  the  number  of  teeth  in 
the  change  gear.  A  constant  dividend  is  a  number  that, 
when  divided  by  the  number  of  teeth  in  a  change  gear, 
gives  the  speed  of  the  driven  shaft  of  a  train  of  gears  and, 
when  divided  by  the  speed  of  the  driven  shaft,  gives  the 
number  of  teeth  in  the  change  gear. 

For  all  speed  calculations  the  constant  number  for  a  train 
of  gears  is  a  constant  factor  if  the  change  gear  is  a  driver 
and  a  constant  dividend  if  the  change  gear  is  a  driven  gear. 
Where  a  constant  number  is  used  in  connection  with  some 
result  dependent  on  the  action  of  a  train  of  gears,  it  may  be 
either  a  constant  factor  or  dividend,  depending  on  whether 
the  value  of  the  said  result  is  increased  or  decreased  by  an 
increase  in  the  size  of  the  change  gear. 

To  find  a  constant: 


Rule. — Perform  the  calculation  of  the  train  of  gears  in  the 
ordinary  manner ,  using  a  theoretical  change  gear  of  one  tooth. 


Example. — A  certain  roll  is  driven  as  follows:  The  first  driving 
gear  has  40  teeth  and  makes  390  revolutions  per  minute;  this  gear 
drives  a  39-tooth  gear  attached  to  a  shaft  on  which  there  is  also  a 
64-tooth  gear  driving  a  32-tooth  gear  on  a  shaft.  On  this  latter  shaft 
is  fastened  a  40-tooth  gear  that  drives  a  40-tooth  gear  on  another  shaft; 
on  this  shaft  a  change  gear  drives  a  128-tooth  gear  on  the  shaft  of  the 
roll.  What  is  the  constant  for  finding  the  speed  of  the  roll  with  various 
change  gears? 

Solution. — In  this  case  the  constant  number  must  be  a  constant 
factor,  as  the  change  gear  is  a  driver  and  an  increase  in  its  size  increases 
the  speed  of  the  roll. 


390  X  40  X  64  X  40  X  1 
39  X  32  X  40  X  128 


6.25,  constant  factor.  Ans. 


Note.— If  any  change  gear  is  used,  its  size  multiplied  by  6.25  in  this  case,  will  give 
the  speed  of  the  roll;  also,  if  any  speed  is  desired,  the  required  change  gear  can 
be  found  by  dividing  the  desired  speed  by  6.25. 


EXAMPLES  FOR  PRACTICE 

1.  The  constant  factor  for  a  certain  train  of  gears  is  7.5;  what 
change  gear  will  give  150  turns  of  the  driven  shaft  per  minute? 

Ans.  20-tooth  gear 


16 


MECHANICAL  CALCULATIONS 


§8 


2.  Find  the  constant  for  the  following  train  of  gears,  the  first  driver 
making  220  revolutions  per  minute:  drivers  24  ,  42,  30,  respectively, 
and  change  gear;  driven  18,  36,  24,  and  42  teeth,  respectively. 

Ans.  10.185,  constant 

3.  The  constant  dividend  for  a  train  of  gears  is  768;  what  will  be 
the  speed  of  the  driven  gear  when  a  24-tooth  change  gear  is  used? 

Ans.  32  rev.  per  min. 

4.  Find  the  constant  for  the  following  train  of  gears  in  which  the 

first  driving  gear  makes  120  revolutions  per  minute:  drivers,  30,  45,  90, 
and  40  teeth,  respectively,  driven,  30,  60,  50,  respectively,  and  change 
gear.  Ans.  6,480,  constant  dividend 


RULES  APPLYING  TO  LEVERS 

17.  To  find  the  weight  supported  or  the  pressure  exerted 
at  the  weight  end  of  a  lever,  the  length  of  the  weight  arm, 
the  length  of  the  power  arm,  and  the  power  applied  being- 
known: 

Rule. — Multiply  the  power  by  the  length  of  the  power  arm 
and  divide  the  product  by  the  length  of  the  zveight  arm. 

Example. — A  25-pound  weight  is  placed  on  a  lever  that  is  so  con¬ 
nected  as  to  exert  a  pressure  on  a  pair  of  rolls;  the  weight  is  4  feet 
from  the  fulcrum  of  the  lever  and  the  rod  connecting  the  lever  with  the 
rolls  is  1|  feet  from  the  fulcrum  of  the  lever;  find  the  pressure  exerted. 

25  X  4 

Solution. —  — — j —  =  66f  lb.  pressure.  Ans. 

18.  Any  problem  of  levers  may  be  solved  by  treating  it 
as  a  proportion  in  which  the  power  is  to  the  weight  as  the 
weight  arm  is  to  the  power  arm;  and,  as  in  proportion 
the  product  of  the  extremes  is  equal  to  the  product  of  the 
means,  so  the  power  times  the  power  arm  is  equal  to 
the  weight  times  the  weight  arm. 

When  dealing  with  compound  levers  it  must  be  remembered 
that  the  continued  product  of  the  power  multiplied  by  the 
power  arms  is  equal  to  the  continued  product  of  the  weight 
multiplied  by  all  the  weight  arms,  every  alternate  arm  of 
the  combination  of  levers,  starting  with  the  power  arm, 
being  a  power  arm  and  every  alternate  arm,  starting  with 
the  weight  arm,  being  a  weight  arm. 


§8 


MECHANICAL  CALCULATIONS 


17 


Example. — A  40-pound  weight  (power)  acts  through  the  following 
power  arms:  4  feet,  3  feet,  and  3  feet,  respectively;  the  corresponding 
weight  arms  being  3  feet,  2  feet,  and  2  feet,  respectively;  what  weight 
is  supported,  or  pressure  exerted,  at  the  extremity  of  the  last 
weight  arm  ? 


40  X  4  X  3  X  3 


120  lb.  Ans. 


Solution. — 


3X2X2 


YARN  CALCULATIONS, 
COTTON 


SINGLE  YARNS 


NUMBERING  SYSTEM 

1.  From  the  earliest  times  of  the  manufacture  of  yarns 
it  has  been  found  necessary  to  adopt  some  system  by  means 
of  which  the  different  sizes  of  yarns  may  be  designated, 
since,  otherwise,  the  manufacturer  would  have  no  means  of 
readily  distinguishing  one  yarn  from  another. 

In  the  cotton  system  of  numbering  yarns,  all  counts  are 
based  on  the  standard  length  of  840  yards,  which  is  known 
as  1  hank;  in  other  words,  a  hank  of  cotton  yarn  contains 
840  yards.  That  this  is  a  constant  number,  never  varying, 
is  a  fact  that  should  always  be  remembered.  For  example, 
1  hank  of  a  certain  number  of  cotton  yarn  contains  840  yards, 
and  a  single  hank  of  any  yarn  either  finer  or  coarser  will 
contain  the  same  number  of  yards. 

The  method  of  numbering  the  yarns  is  that  of  calling  a 
yarn  that  contains  1  hank,  or  840  yards,  in  1  pound  a  No.  1 
yarn.  If  the  yarn  contains  2  hanks,  or  1,680  yards  in 
1  pound,  it  is  known  as  a  No.  2  yarn;  if  it  contains  3  hanks, 
or  2,520  yards,  in  1  pound,  it  is  known  as  a  No.  3  yarn.  Thus 
it  will  be  seen  that  the  number  of  hanks  (840  yards)  that  it 
takes  to  weigh  1  pound  determines  the  counts  of  the  yarn. 

The  counts  of  a  yarn  are  generally  indicated  by  placing  a 
letter  s  after  the  figure  representing  the  number  of  the  yarn. 
Thus,  26s  shows  the  counts  of  a  yarn  and  indicates  that  the 
yarn  contains  26  hanks  (26  X  840  yards)  in  1  pound. 


For  notice  of  copyright,  see  page  immediately  following  the  title  page 

§9 


2 


YARN  CALCULATIONS,  COTTON 


§9 


CALCULATIONS  OF  SINGLE  YARNS 

2.  There  are  several  calculations  that  become  necessary 
when  dealing  with  single  yarns,  but  all  are  based  on  one 
equation,  namely, 

length  of  yarn,  in  yards  _  ^ 

weight,  in  pounds  X  counts  X  standard 

The  standard  will  always  be  the  same  when  dealing  with 
cotton  (840  yards),  and  any  one  of  the  other  items  can  be 
found  by  simply  substituting  the  values  of  the  rest  in  the 
above  equation. 

3.  To  find  the  counts  of  a  yarn  when  the  length  and 
weight  are  given: 

Rule. — Divide  the  total  length  of  yarn ,  expressed  in  yards, 
by  the  weight ,  expressed  in  pounds ,  times  the  standard  length. 

Example. — If  168,000  yards  of  yarn  weighs  5  pounds,  what  are 
the  counts? 

Solution. — 

168,000  (length  of  yarn,  in  yards) 

■=—, — r-rr  — ,  .  w  Q ,  , — y — yv  =  40s,  counts.  Ans. 

5  (weight,  in  pounds)  X  840  (standard) 


4.  To  find  the  weight  of  yarn  when  the  length  and 
counts  are  known: 


Rule. — Divide  the  length ,  in  yards,  by  the  counts  times  the 
standard  length. 


Example. — What  is  the  weight  of  42,000  yards  of  number  5s  yarn? 
42,000  (length,  in  yards) 


Solution. - 


5  (counts)  X  840  (standard) 


=  10  lb.  Ans. 


5.  To  find  the  length  of  yarn  when  the  weight  and 
counts  are  known: 

Rule. — Multiply  the  weight,  in  pou?ids,  counts,  a?id  standard 
length  together. 

Example. — What  is  the  length  of  yarn  contained  in  a  bundle  that 
weighs  8  pounds,  the  counts  of  the  yarn  being  26s? 

Solution. —  8  (weight,  in  pounds)  X  26  (counts)  X  840  (standard) 
=  174,720  yd.  Ans. 


$9 


YARN  CALCULATIONS,  COTTON 


3 


6.  When  numbering  yarns  several  different  weights  and 
names  are  used,  for  which  reason  it  will  be  well  to  memorize 
the  two  following  tables.  These  tables  include  only  those 
terms  used  in  yarn  numbering. 

TABLE  I 

The  table  of  lengths  is  as  follows: 

li  yards  =  1  thread,  or  circumference  of  wrap  reel. 

120  yards  =  80  threads  =  1  skein,  or  lea. 

840  yards  =  -560  threads  =  7  skeins,  or  leas  =  1  hank. 

TABLE  II 

The  table  of  weights  is  as  follows: 

27.34  grains  =  1  dram. 

437.5  grains  =  16  drams  =  1  ounce. 

7,000  grains  =  256  drams  =  16  ounces  =  1  pound. 

The  part  of  this  table  that  is  most  frequently  used  and 
which,  therefore,  should  be  carefully  noted  is  that  which 
states  that  there  are  7,000  grains  in  1  pound. 


SIZING  ROVING  AND  YARN 

7.  From  the  time  the  cotton  is  placed  in  the  form  of 
roving  until  it  becomes  the  finished  yarn,  some  method  has 
to  be  adopted  in  every  well-regulated  mill  by  means  of  which 
the  exact  counts  of  the  roving  and  yarn  being  run  may  be 
known. 

Roving  is  a  term  used  to  designate  a  strand  of  loosely 
twisted  fibers  from  which  a  yarn  may  be  spun,  while  the 
term  yai*n  is  applied  to  a  thread  composed  of  fibers  uniformly 
disposed  throughout  its  structure  and  having  a  certain 
amount  of  twist  for  the  purpose  of  enhancing  its  strength. 

It  is  generally  the  custom  to  weigh  a  certain  quantity  from 
each  machine,  at  least  once  a  day,  and  by  this  means  ascer¬ 
tain  whether  the  roving  or  yarn  is  being  kept  at  the 
required  weight.  This  process  is  known  as  sizing,  and  is 
a  matter  that  should  always  be  carefully  attended  to. 

From  the  rules  and  explanations  previously  given  it  will 
be  plain  that  if  840  yards  (1  hank)  was  always  the  length 


4  YARN  CALCULATIONS,  COTTON  §9 

weighed,  in  order  to  learn  the  counts  of  the  yarn,  it  would 
simply  be  necessary  to  divide  the  weight,  expressed  in 


Fig.  1 


pounds,  into  1  pound,  or  if  expressed  in  grains,  into  7,000 
(the  number  of  grains  in  1  pound).  It  will  readily  be  seen 


Fig.  2 

that  to  measure  off  840  yards  of  yarn  would  not  only  require 
considerable  time,  but  would  also  produce  an  unnecessary 


§9 


YARN  CALCULATIONS,  COTTON 


5 


waste  of  material.  To  overcome  these  difficulties,  when 
sizing  yarn,  it  is  customary  to  measure  off  one-seventh  of 
840  yards,  or  120  yards;  to  weigh  this  amount;  and  divide  its 
weight,  in  grains,  into  one-seventh  of  7,000,  or  1,000.  The 
result  obtained  in  this  manner  will  be  the  same  as  if  840  yards 
was  taken  and  the  weight,  in  grains,  divided  into  7,000. 

8.  The  Wrap  Reel. — When  sizing  yarns,  an  instrument 
known  as  a  wrap  reel  is  used  to  measure  the  yarn.  As  its 
name  indicates,  this  instrument  consists  of  a  reel,  generally 
I2  yards  in  circumference.  The  yarn  is  wound  on  this 
reel  and  a  finger  indicates  on  a  disk  the  number  of  yards 
reeled.  Fig.  1  shows  an  ordinary  type  of  wrap  reel,  while 
Fig.  2  shows  yarn  and  roving  scales.  These  scales  are  suit¬ 
able  for  weighing  by  tenths  of  grains. 

Example. —  120  yards  of  yarn  is  reeled  and  found  to  weigh 

40  grains;  what  are  the  counts? 

Solution. —  1,000  -p  40  =  25s.  Ans. 

9.  This  basis  for  indicating  the  size  of  the  yarn  is  also 
used  to  designate  the  size  of  roving,  although  when  sizing 
roving  a  shorter  length  is  used.  It  is  customary  in  this  case 
to  measure  off  one-seventieth  of  840  yards,  or  12  yards,  and 
divide  the  weight,  in  grains,  of  this  length  of  roving  into 
one-seventieth  of  7,000,  or  100. 

Example. —  12  yards  of  roving  is  found  to  weigh  20  grains;  what 
are  the  counts? 

Solution. —  100  -p  20  =  5-hank.  Ans. 

The  counts  of  roving  are  indicated  in  a  somewhat  different 
manner  from  the  counts  of  yarn.  Thus,  in  this  case,  the 
roving  would  be  known  as  5-hank  ?vving,  indicating  that 
5  hanks  weigh  1  pound. 

10.  From  the  foregoing,  it  will  be  readily  understood 
that  the  counts  of  a  yarn  may  be  obtained  by  finding  the 
weight  of  any  length  of  yarn  and  dividing  the  weight  into  a 
certain  number  of  grains,  this  number  having  the  same  pro¬ 
portion  to  7,000  that  the  length  of  yarn  weighed  has  to  840. 


6 


YARN  CALCULATIONS,  COTTON 


9 


As  an  aid  to  the  student  in  any  calculations  that  he  may 
have  to  deal  with,  the  following  lengths  and  dividends 
are  given: 

TABLE  III 


If  480  yards  is  weighed,  divide 
If  240  yards  is  weighed,  divide 
If  120  yards  is  weighed,  divide 
If  60  yards  is  weighed,  divide 
If  40  yards  is  weighed,  divide 
If  30  yards  is  weighed,  divide 
If  20  yards  is  weighed,  divide 
If  15  yards  is  weighed,  divide 
If  12  yards  is  weighed,  divide 
If  10  yards  is  weighed,  divide 
If  8  yards  is  weighed,  divide 
If  6  yards  is  weighed,  divide 
If  4  yards  is  weighed,  divide 
If  3  yards  is  weighed,  divide 
If  2  yards  is  weighed,  divide 
If  1  yard  is  weighed,  divide 


weight  in  grains  into  4,000. 
weight  in  grains  into  2,000,/ 
weight  in  grains  into  1,000. 
weight  in  grains  into  500. 
weight  in  grains  into  333# . 
weight  in  grains  into  250. 
weight  in  grains  into  166f. 
weight  in  grains  into  125. 
weight  in  grains  into  100. 
weight  in  grains  into  83#. 
weight  in  grains  into  66#. 
weight  in  grains  into  50. 
weight  in  grains  into  33#. 
weight  in  grains  into  25. 
weight  in  grains  into  16#. 
weight  in  grains  into  8#. 


EXAMPLES  FOR  PRACTICE 

1 .  50,400  yards  of  cotton  yarn  weighs  3  pounds;  what  are  the  counts? 

Ans.  20s 

2.  100,800  yards  of  cotton  yarn  weighs  10  pounds;  what  are  the 

counts?  Ans.  12s 

3.  100  skeins  of  cotton  yarn,  each  containing  4,200  yards,  weighs 

20  pounds;  what  are  the  counts?  Ans.  25s 


4.  What  are  the  counts  of  cotton  yarns  120  yards  of  which  weighs, 


respectively,  17,  21,  26,  and  32  grains?  * 


Ans. 


58.82s 

47.61s 

38.46s 

131.25s 


5.  If  a  cop  is  known  to  contain  960  yards  of  cotton  yarn  and  35  of 

these  cops  weigh  1  pound,  what  are  the  counts?  Ans.  40s 

6.  What  is  the  weight  of  37,000  yards  of  Is  cotton?  Ans.  44.04  lb. 

7.  What  is  the  weight,  in  ounces,  of  10,000  yards  of  36s  cotton? 

Ans.  5.29  oz.  (practically) 

8.  How  manyyards  of  60s  cotton  in  24  pounds?  Ans.  1,209,600  yd. 


§9 


YARN  CALCULATIONS,  COTTON 


7 


9.  What  is  the  length  of  cotton  yarn  in  50  pounds  of  20s? 

Ans.  840,000  yd. 

10.  A  spool  of  28s  cotton  yarn  weighs  28  ounces.  Find  the  length 
of  yarn  on  the  spool  if  the  spool  itself  weighs  8  ounces.  Ans.  29,400  yd. 

11.  What  is  the  weight  of  336,000  yards  of  40s  cotton?  Ans.  10  lb. 

12.  What  is  the  weight  of  151,200  yards  of  60s  cotton?  Ans.  3  lb. 


PLY  YARNS 


METHOD  OF  NUMBERING 

11.  Definition. — Very  frequently  during  the  process  of 
manufacturing  yarns,  it  becomes  necessary  to  twist  two  or 
more  threads  together  to  form  one  coarser  thread.  Such 
yarns  are  commonly  known  as  ply  yarns,  also  sometimes 
called  folded,  or  twisted,  yarns.  It  should  be  understood 
that  when  yarns  are  folded  in  this  manner  it  does  not  change 
their  length,  with  the  exception  of  a  slight  percentage  of 
contraction,  which  takes  place  during  the  twisting,  but  it  does 
change  their  weight. 

The  method  of  numbering  cotton  ply  yarns  is  that  of 
giving  the  counts  of  the  single  yarns  that  are  folded  and 
placing  before  these  counts  the  number  that  indicates  the 
number  of  threads  folded;  thus,  2/40s  indicates  that  two 
threads  of  40s  single  yarn  are  folded  together,  causing  the 
ply  yarn  to  be  equal  in  weight  to  a  single  20s.  As  previ¬ 
ously  stated,  during  the  process  of  twisting,  a  slight  con¬ 
traction  takes  place.  Consequently,  to  make  the  resultant 
counts  20s,  the  single  yarns  folded  would  have  to  be  slightly 
finer  than  40s.  However,  this  contraction  will  not  be  con¬ 
sidered  in  the  rules  and  examples  that  follow,  since  it  is  so 
slight  that  it  is  more  a  matter  of  experience  than  one  of 
mathematics. 


8 


YARN  CALCULATIONS,  COTTON 


§9 


CALCULATIONS  OF  PLY  YARNS 


FOLDED  YARNS  OF  THE  SAME  COUNTS 


12.  It  is  not  often  the  custom  in  mills  to  fold  yarns  of 
different  counts,  since  single  yarns  of  the  same  number 
make  the  best  double,  or  ply,  yarns.  Consequently,  when 
yarns  of  the  same  counts  are  folded,  in  order  to  find  the 
counts  of  the  resulting  ply  yarn,  it  is  simply  necessary  to 
divide  the  counts  of  the  yarns  folded  by  the  number  of 
threads  that  constitute  the  ply  yarn.  For  example,  if  three 
threads  of  90s  cotton  are  folded  to  form  a  ply  yarn,  the 
resultant  yarn  will  be  equivalent  in  weight  to  a  single  30s 
(90  4-  3  =  30). 

The  method  of  indicating  the  counts  of  this  ply  yarn  is 
3/90s.  The  rules  for  finding  the  counts,  weight,  and  length 
of  ply  yarns  are  similar  to  those  given  for  finding  the  same 
particulars  of  single  yarns,  with  the  exception  that  the 
counts  of  the  ply  yarn  do  not  indicate  the  actual  counts  of 
the  yarn,  but  instead  indicate  the  counts  of  the  yarns  folded. 
Consequently,  when  figuring  to  find  these  particulars,  the 
actual  counts  of  the  folded  yarn  must  be  taken.  The 
following  examples  will  make  this  subject  clearer: 


Example  1. — What  is  the  weight  of  642,000  yards  of  2-ply  40s 
cotton  yarn? 


Solution. — 


2-ply  40s  cotton  yarn  is  equal  in  weight  to  a  single  20s. 
642,000 


20  X  840 


=  38.21  lb.  Ans. 


Example  2. — What  is  the  length  of  20  pounds  of  2-ply  60s  cotton? 
Solution. — The  2-ply  60s  cotton  is  equal  in  weight  to  a  single  30s. 
20  X  30  X  840  =  504,000  yd.  Ans. 


Example  3. — What  are  the  counts  of  a  2-ply  yarn  352,800  yards  of 
which  weighs  10  pounds? 


Solution. — 


_352,800 
10  X  840 


=  42s 


This  answer  is  the  counts  of  the  ply  yarn  considered  as  a  single 
thread,  but  since  it  is  made  from  two  threads  of  the  same  counts  it  will 
be  a  2-ply  84s.  Therefore,  352,800  yd.  of  the  2-ply  84s  cotton  yarn 
will  weigh  10  lb.  Ans. 


§9 


YARN  CALCULATIONS,  COTTON 


9 


FOLDED  YARNS  OF  DIFFERENT  COUNTS 

13.  As  was  previously  stated,  it  does  not  often  occur  that 
different  counts  are  folded  together  to  form  a  ply  yarn,  since 
singles  of  the  same  number  make  the  best  double,  or  ply, 
yarns.  However,  this  will  sometimes  happen,  especially 
when  it  is  desired  to  make  a  fancy  yarn,  in  which  case 
two  yarns  of  different  counts  are  often  folded. 

For  an  illustration,  suppose  that  it  is  desired  to  find  the 
resultant  counts  of  a  40s  folded  with  a  20s.  Take  as  a  basis 
840  yards  of  each  yarn;  then  840  yards  of  the  40s  weighs 
tV  pound.  840  yards  of  the  20s  weighs  2V  pound.  Conse¬ 
quently,  after  these  yarns  are  folded  there  will  be  840  yards 
of  a  ply  yarn  the  weight  of  which  is  i1?  +  2V  =  A  pound. 
The  example  now  resolves  itself  into  the  following:  What 
are  the  counts  of  a  yarn  840  yards  of  which  weighs  impound? 
Since  length  divided  by  (weight  times  standard)  equals 

counts,  then  -3  — ^0— -  =  13.33s,  counts  of  the  ply  yarn. 

To  X  840 

This  example  has  been  worked  out  to  some  length  in 
order  to  enable  the  student  to  thoroughly  understand  the 
method  of  numbering  ply  yarns.  A  somewhat  shorter 
method  is  as  follows: 

14.  To  find  the  resultant  counts  when  two  threads  of 
different  numbers  are  folded: 

Rule. — Multiply  the  tzvo  counts  together  and  divide  the  result 
thus  obtained  by  the  sum  of  the  counts. 

Example. — What  will  be  the  counts  of  a  yarn  obtained  by  folding 
a  40s  with  a  20s? 

Solution. —  ^  =  13.33s,  counts.  Ans. 

40  -f  20 


PLY  YARNS  COMPOSED  OF  MORE  THAN  TWO  THREADS 

15.  In  many  cases  it  will  be  found  necessary  to  find  the 
counts  of  a  ply  yarn  that  is  made  from  more  than  two  single 
threads.  Under  such  circumstances  a  somewhat  different 
process  must  be  followed.  For  example,  suppose  that 


10 


YARN  CALCULATIONS,  COTTON 


9 


one  thread  each  of  24s,  36s,  and  72s  are  folded  to  form  a 
ply  yarn  and  it  is  required  to  ascertain  the  counts  of  the 
resultant  yarn.  This  may  be  found  by  following  the  rule 
given  and  performing  two  operations  as  follows: 

First  find  the  counts  of  the  yarn  that  will  result  from  fold¬ 
ing  the  24s  with  the  36s.  —  *  --  =  14.4s.  The  example 

24  +  36 

then  resolves  itself  into  the  following:  What  are  the  counts 
of  a  ply  yarn  made  from  one  thread  of  14.4s  and  one  of  72s. 


14.4  X  72 
14.4  +  72 


12s.  Ans. 


16.  A  somewhat  shorter  method  than  this,  however,  may 
be  applied  to  3  or  more  ply  yarns  made  from  different  counts. 


Rule.  —  Take  the  highest  counts  and  divide  it  by  itself  and  by 
each  of  the  other  counts.  Add  the  results  thus  obtamed  a?id 
divide  this  result  into  the  highest  counts. 

Example. — Same  as  in  the  preceding  article. 

Solution. —  72  -r-  72  =  1 

72  -I-  36  =  2 
72  -I-  24  =  3 

6 

72  -I-  6  =  12s.  Ans. 


17.  To  find  the  resultant  counts  when  more  than  one 
end  of  the  different  counts  are  folded: 

Rule. — Divide  the  highest  counts  by  itself  and  by  each  of  the 
other  counts.  Multiply  the  result  in  each  case  by  the  number  of 
ends  of  that  counts.  Add  the  results  thus  obtained  and  divide 
this  result  into  the  highest  counts. 

Example. —  4  ends  of  80s  and  3  ends  of  60s  are  folded  to  form  a 
ply  yarn;  what  are  the  resultant  counts? 

Solution. —  80  =  80  =  1;  1X4  =  4 

80  =  60  =  H;  H  X  3  =  4 

8 

80  -7-  8  =  10s,  resultant  counts.  Ans. 

18.  Another  calculation  that  will  frequently  occur  when 
dealing  with  ply  yarns  is  that  of  finding  the  counts  of  a  yarn 
that  must  be  folded  with  another  to  produce  the  given  counts. 


§9 


YARN  CALCULATIONS,  COTTON 


11 


Rule. —  To  find  what  counts  must  be  folded  with  another  to 
produce  a  given  counts,  nniltiply  the  two  counts  together  and 
divide  by  their  difference. 


Example. — What  counts  of  yarn  must  be  folded  with  a  50s  to  pro¬ 
duce  a  30s  ply  yarn? 


Solution. — 


50  X  30 
50  -  30 


75s.  Ans. 


The  student  may  prove  this  example  by  taking  the  two 
single  yarns,  50s  and  75s,  and  finding  the  ply  yarn  resulting 
from  folding  them,  following  the  rule  given  for  finding  the 
counts  of  ply  yarn  resulting  from  the  folding  of  single  yarns. 


19.  Still  another  calculation  that  enters  into  ply  yarns  is 
that  of  finding  the  weight  of  single  yarns  of  different  counts 
that  are  to  be  twisted  together  to  produce  a  given  weight  of 
ply  yarn. 

Rule. —  When  tzvisting  together  two  or  more  threads  of  dif¬ 
ferent  counts ,  to  find  the  weight  of  each  required  to  produce 
the  giveti  weight ,  find  the  counts  resulting  from  folding  the 
two  or  more  threads;  then  as  the  counts  of  one  thread  is  to 
the  resulting  counts  so  is  the  total  weight  to  the  weight  required 
of  that  thread. 

Example. — It  is  desired  to  produce  100  pounds  of  a  ply  yarn  from 
an  80s  and  a  32s;  what  weight  of  each  single  thread  must  be  used  in 
order  to  produce  the  100  pounds  of  ply  yarn? 


Solution. — 


80  X  32 
80  +  32 


22.85s 


32  :  22.85  =  100  :  71.40 


Therefore,  71.40  lb.  of  32s  would  be  required. 


Ans. 


It  will  readily  be  seen  that  in  order  to  find  the  weight  of  the  other 
thread  it  is  only  necessary  to  subtract  the  weight  of  the  32s  from  the 
total  weight,  or  100  lb.  But  in  order  that  this  may  be  fully  under¬ 
stood,  the  required  weight  of  80s  will  be  worked  out  similarly  to  that 
of  32s. 

Resulting  counts  =  22.85;  80  :  22.85  =  100  :  28.56  lb.  of  80s.  Ans. 


Note.— In  the  previous  example  the  weight  of  the  80s  yarn  plus  the  weight  of 
the  32s  yarn  should  equal  the  weight  of  the  ply  yarn,  but  owing  to  the  use  of  decimals, 
examples  of  this  kind  seldom  give  exact  results.  Thus,  71.40  pounds  +  28.56  pounds. 
=  99.96  pounds;  whereas  the  total  weight  should  be  100  pounds. 


12 


YARN  CALCULATIONS,  COTTON 


§9 


exampi.es  for  practice 

1.  If  a  thread  of  40s  and  one  of  60s  are  twisted  together,  what  are 

the  counts  of  the  resultant  ply  yarn?  Ans.  24s 

2.  A  20s,  30s,  and  60s  are  twisted  together;  what  are  the  counts  of 

the  ply  yarn?  Ans.  10s 

3.  What  are  the  counts  of  a  yarn  that  must  be  twisted  with  a  44s 

to  produce  a  ply  yarn  equal  to  a  20s?  Ans.  36.66s 

4.  What  weight  of  40s  must  be  twisted  with  a  20s  to  produce 

200  pounds  of  ply  yarn?  Ans.  66.65  lb. 

5.  What  weight  of  100s,  80s,  and  60s  would  be  used  in  producing 

500  pounds  of  ply  yarn?  r  127.65  lb.  of  100s 

Ans.  {  159.56  lb.  of  80s 
1212.75  lb.  of  60s 


6.  What  counts  of  ply  yarn  will  result  if  18s,  24s,  and  36s  are 
twisted  together,  and  what  weight  of  each  will  there  be  in  240  pounds 


of  the  ply  yarn? 


Ans. 


8s 

106|  lb.  of  18s 
80  lb.  of  24s 
53i  lb.  of  36s 


7.  A  ply  yarn  is  made  from  1  end  each  of  10s,  12s,  and  15s;  what 

are  the  counts  of  the  ply  yarn?  Ans.  4s 

8.  A  3-ply  yarn  is  made  from  80s,  40s,  and  30s  and  weighs 
100  pounds;  what  weight  does  it  contain  of  each  counts  of  single  yarn 


and  what  are  the  counts  of  the  ply  yarn? 


Ans. 


14.117s 

17.64  lb.  of  80s 
35.29  lb.  of  40s 
147.05  lb.  of  30s 


9.  What  will  be  the  resultant  counts  if  2  ends  of  40s  and  1  end  of 

60s  are  twisted  to  make  a  ply  yarn?  Ans.  15s 

10.  What  counts  result  from  twisting  a  60s  with  a  36s?  Ans.  22.5s 


9 


YARN  CALCULATIONS,  COTTON 


13 


BEAM  CALCULATIONS 

20.  The  yarn  that  forms  the  warp  of  a  piece  of  cloth 
after  being  spun  is  placed  on  what  are  known  as  section 
beams,  and  is  then  run  from  these  section  beams  on  to  the 
loom  beam,  which  is  placed  at  the  back  of  the  loom.  The 
yarn  is  then  ready  to  be  made  into  cloth.  Whether  this 
warp  yarn  is  on  the  section  beam  or  loom  beam,  there  are 
several  calculations  that  it  is  necessary  to  understand  and 
that  apply  equally  well  to  both  cases. 

21.  To  find  the  counts  of  the  yarn  on  a  beam  that  con¬ 
tains  one  size  of  yarn,  the  weight,  length,  and  number  of 
ends  being  given: 

Rule. — Multiply  the  length ,  in  yards ,  by  the  mimber  of  ends 
on  the  beam  and  divide  the  result  thus  obtained  by  the  weight , 
in  pounds,  times  the  standard  mimber  of  yards  to  the  pound. 

Example. — A  warp  beam  contains  2,400  ends  of  cotton,  each 
200  yards  long;  the  weight  of  this  yarn  is  15  pounds.  What  are 
the  counts? 

c  200  X  2,400  OQ  . 

Solution. —  --rz - _ =  38.095s.  Ans. 

15  X  840 

Note.— It  has  been  shown  how  it  is  possible  to  obtain  the  counts  of  a  yarn  when 
its  weight  and  length  are  given.  The  same  rule  is  adopted  in  this  case  with  the 
exception  that  the  length  of  only  one  end  is  given  and,  as  there  are  several  on  a 
beam,  it  is  necessary  to  multiply  the  length  of  each  end  by  the  number  of  ends  on  the 
beam  in  order  to  obtain  the  total  length  of  yarn. 

It  should  be  carefully  noted  in  connection  with  these 
examples  that  in  many  cases  the  weight  given  will  be  found 
to  include  not  only  the  weight  of  the  yarn,  but  also  that  of 
the  beam  on  which  it  is  placed.  Under  such  circumstances 
it  becomes  necessary  to  deduct  the  weight  of  the  beam  from 
the  weight  given,  in  order  to  find  the  true  weight  of  the  yarn. 

22.  To  find  the  number  of  ends  on  a  beam  when  the 
weight,  length,  and  counts  are  given: 


14 


YARN  CALCULATIONS,  COTTON 


§9 


Rule. — Multiply  the  weight,  in  pounds,  by  the  standard  num¬ 
ber  of  yards  and  also  by  the  counts.  Divide  this  result  by  the 
length  of  the  warp,  in  yards. 


Example. — The  yarn  on  a  beam  weighs  200  pounds;  the  counts  are 
20s  cotton,  and  the  length  of  each  end  1,400  yards.  How  many  ends 
does  the  beam  contain? 


Solution. — 


200  X  840  X  20 
1,400 


2,400  ends.  Ans. 


23.  To  find  the  weight  of  yarn  on  a  beam  when  the 
length,  number  of  ends,  and  counts  are  given: 


Rule. — Multiply  the  length,  in  yards,  by  the  number  of  ends 
on  the  beam  and  divide  the  result  thus  obtained  by  the  standard 
number  of  yards  times  the  counts  of  the  yarn. 

Example. — A  beam  contains  2,400  ends  of  20s  cotton,  each  end 
being  500  yards  long;  find  the  weight  of  the  yarn. 

_  500  X  2,400 

Solution. —  —  =  71.428  lb.  Ans. 

o4U  X 


24.  To  find  the  length  of  warp  on  a  beam  when  the 
weight,  number  of  ends,  and  counts  are  given: 

Rule. — Multiply  the  weight,  in  pounds,  by  the  standard 
length  and  by  the  counts.  Divide  the  result  thus  obtained  by 
the  number  of  ends  on  the  beam. 


Example. — It  is  required  to  make  a  warp  of  1,600  ends  of  20s  from 
30  pounds  of  raw  stock;  what  will  be  the  length  of  the  warp? 


Solution. — 


30  X  840  X  20 
1,600 


315  yd.  Ans. 


25.  To  find  the  length  of  warp  that  can  be  placed  on 
a  beam: 


Rule. — Find  the  weight  of  yarn  that  the  beam  will  contain, 
by  weighing  a  beam  of  the  same  size  when  filled  with  yarn  and 
deducting  the  weight  of  the  beam  itself.  Then  apply  the  rule 
in  Art.  24. 

Example  1. — A  certain  size  beam  when  filled  with  yarn  weighs 
140  pounds,  the  beam  itself  weighing  50  pounds.  What  length  of  a 
warp  composed  of  1,800  ends  of  20s  cotton  can  be  placed  on  it? 


§9 


YARN  CALCULATIONS,  COTTON 


15 


Solution. — 


140  —  50  =  90  lb.  of  yarn. 
90  X  840  X  20 


1,800 


=  840  yd. 


Ans. 


Example  2. — Same  beam  as  in  example  1.  How  many  yards  of  a 
warp  containing  3,000  ends  of  40s  cotton  can  be  placed  on  it? 


Solution.— 


140  —  50  =  90  lb.  of  yarn. 


90  X  840  X  40 
3,000 


1,008  yd. 


Ans. 


AVERAGE  NUMBERS 

26.  Very  frequently  different  counts  of  yarn  will  be 
found  on  the  same  beam  or  in  the  same  warp.  When  this 
is  the  case,  it  becomes  necessary  to  find  the  average  number , 
or  average  counts ,  of  the  different  yarns  before  other  calcula¬ 
tions  can  be  made.  By  the  term  average  number,  or 
average  counts,  of  tlie  warp,  a  count  of  yarn  is  meant 
that  will  give  the  same  weight  as  the  warp  yarn,  considering 
that  an  equal  number  of  ends  are  used. 

27.  To  find  the  average  number  of  the  warp  yarn  when 
more  than  one  count  is  used: 

Rule. — Divide  the  total  number  of  ends  of  each  count  by  its 
own  count.  Add  these  results  together  and  divide  the  result  thus 
obtained  into  the  total  number  of  ends  in  the  warp. 

Example. — Suppose  that  on  the  same  beam  there  are  1,800  ends 
of  60s  and  800  ends  of  40s;  what  counts  of  yarn  will  weigh  the  same  as 
the  total  weight  of  warp  yarn,  using  the  same  number  of  ends  in 
both  cases? 

Solution. —  1800-5-60  =  30 

8  0  0  -s-  40  =  20 

2600  50 

2,600  -T-  50  =  52s,  average  counts.  Ans. 

Proof. — That  the  above  example  is  correct  can  be  shown 
by  finding  the  weight  of  the  warp  yarn  and  also  the  weight 
of  2,600  ends  of  52s,  considering  the  length  to  be  1  yard. 

Applying  the  rule  given  for  finding  weight  when  ends, 

length,  and  counts  are  g.ven,  —  — ^  =  .0357  lb., 


16 


YARN  CALCULATIONS,  COTTON 


9 


=  .0238  lb.;  .0357  +  .0238  =  .0595  lb.  =  the  total  weight 
of  1  yard  of  warp  yarn. 

Applying  the  same  rule  for  finding  the  weight  of  2,600 

ends  of  52s,  X  2,600  _  ^  Ans. 

840  X  52 

28.  The  above  rule  will  be  found  to  apply  equally  well 
when  more  than  two  counts  are  used  in  the  same  warp. 

Example. — What  will  be  the  average  number  of  a  warp  that  con¬ 
tains  200  ends  of  20s,  1,000  ends  of  40s,  and  900  ends  of  45s? 

Solution. — Applying  the  rule  just  given, 

200  -v-  20  =  10 

1,000  -p  40  =  25 
900  -v-  45  =  2  0 

5  5 

Total  ends  =  2,100;  2,100  -f-  55  =  38.18s,  average  number.  Ans. 

29.  In  cases  where  the  order  of  arranging  the  different 
counts  of  yarn  in  the  warp  is  known,  but  the  total  number  of 
ends  is  not  given,  the  same  rule  will  be  found  to  apply  if  the 
number  of  ends  in  the  arrangement  is  considered  as  the  total 
number  of  ends. 

Example. — A  warp  is  arranged  48  ends  of  36s  and  2  ends  of  10s. 
Find  the  average  number. 

Solution. —  48  -f-  36  =  1.3  3  3 

2  -s-  10  =  .2 

1.5  3  3 

Total  number  of  ends  =  50;  50  1.533  =  32.615s,  average  number. 

Ans. 


FANCY  WARPS 

30.  When  more  than  one  color  of  yarn  is  placed  on  the 
same  beam,  it  frequently  becomes  necessary  to  find  the  total 
number  of  ends  of  each  color,  and  also  the  weight  of  each 
particular  yarn. 

31.  To  find  the  number  of  ends  of  each  color  of  yarn  on 
a  beam  when  the  total  number  of  ends  are  given: 

Rule. — Multiply  the  number  of  ends  of  any  one  color  in  one 
pattern  by  the  total  number  of  ends  on  the  beam  and  divide  by 
the  total  ends  in  one  pattern. 


9 


YARN  CALCULATIONS,  COTTON 


17 


Note. — One  repeat  of  the  colors  in  a  warp  is  termed  the  pattern 
of  tlie  warp.  For  an  illustration,  suppose  that  the  ends  on  a  beam 
are  arrauged  16  ends  black,  8  ends  white,  16  ends  black,  8  ends  gray. 
This  is  known  as  the  pattern  of  the  warp,  since  all  the  ends  are  only 
repeats  of  this  arrangement. 

Example. — Suppose  there  are  2,400  ends  on  a  beam  arranged  as 
stated  in  the  note;  how  many  ends  of  each  color  will  there  be? 

Solution. —  16  ends  black,  8  ends  white,  16  ends  black,  and 
8  ends  gray,  give  48  as  the  total  number  of  ends  in  1  pattern. 

There  are  32  ends  of  black  in  1  pattern. 

32  X  2,400  =  76,800;  76,800  -t-  48  =  1,600  ends  of  black  on  the  beam. 

Aus. 

There  are  8  ends  of  white  in  1  pattern. 

8  X  2,400  =  19,200;  19,200  h-  48  =  400  ends  of  white  on  the  beam. 

Ans. 

There  are  8  ends  of  gray  in  1  pattern. 

8  X  2,400  =  19,200;  19,200  -s-  48  =  400  ends  of  gray  on  the  beam. 

Ans. 

After  the  total  number  of  ends  of  each  color  is  found  in 
this  manner,  in  order  to  learn  the  weight  of  each  color  of  yarn 
it  is  simply  necessary  to  apply  the  rule  given  for  finding  the 
weight  when  the  length,  counts,  and  number  of  ends  are 
given. 


EXAMPLES  FOR  PRACTICE 

1.  What  is  the  weight  of  a  warp  200  yards  long,  that  contains  2,400 

ends  of  60s  cotton?  Ans.  9.52  lb. 

2.  What  are  the  average  counts  in  a  warp  made  from  24  ends  of 

40s  and  1  end  of  8s  cotton?  Ans.  34.48s 

3.  A  warp  is  arranged  3  ends  black;  1,  white;  7,  black;  1,  white; 

counts,  24s  cotton;  154  yards  long;  2,400  ends;  find  the  weight  of  each 
color  and  the  total  weight  of  the  warp.  ( 15.27  lb.  of  black 

Ans.  <3.05  lb.  of  white 

1 18.32  lb. ,  total  weight 

4.  A  warp  is  arranged  24  ends  of  48s  and  1  end  of  2-ply  10s  cotton; 

what  are  the  average  counts?  Ans.  35.71s 

5.  What  is  the  weight  of  warp  yarn  in  a  warp  400  yards  long, 

that  contains  3,360  ends  of  40s  cotton?  Ans.  40  lb. 


6.  4  pounds  of  25s  cotton  is  made  into  a  warp  containing  875  ends; 

what  is  the  length?  Ans.  96  yd. 


18 


YARN  CALCULATIONS,  COTTON 


§9 


7.  What  is  the  length  of  a  warp  made  of  30s  cotton  and  con¬ 
taining  1,800  ends,  if  the  weight  is  1  pound?  Ans.  14  yd. 

8.  A  warp  is  arranged  3  ends  black,  1  end  white,  1  black,  3  white, 

and  weighs  9  pounds;  what  is  the  weight  of  each  kind  of  yarn  if  the 
counts  are  36s  cotton?  Ans.  4.5  lb. 

9.  What  are  the  counts  of  warp  yarn  in  a  cotton  warp  200  yards 
long  that  contains  2,400  ends,  and  weighs  15  pounds?  Ans.  38.09s 

10.  A  cotton  warp  120  yards  long  contains  4,480  ends,  and  weighs 

40  pounds;  what  are  the  counts?  Ans.  16s 

11.  How  many  ends  are  there  in  a  warp  1,400  yards  long  and 
weighing  200  pounds,  the  counts  of  the  yarn  being  20s  cotton? 

Ans.  2,400  ends 

12.  What  will  be  the  cost  of  a  warp  made  of  30s  cotton  at  16  cents 
per  pound,  if  the  warp  is  500  yards  long,  and  contains  1,800  ends? 

Ans.  $5.71 

13.  How  many  ends  are  there  in  a  warp  700  yards  long  and 
weighing  100  pounds,  the  counts  being  40s  cotton?  Ans.  4,800  ends 


TWIST  IN  YARNS 

32.  To  enable  the  yarn,  whether  warp  or  filling,  to  with¬ 
stand  the  strain  that  is  brought  on  it,  it  is  necessary  to  insert 
what  is  known  as  twist.  Warp  yarn  contains  more  twist 
per  inch  than  filling  yarn,  since  it  has  to  withstand  a  greater 
strain  during  the  process  of  weaving.  The  common  practice 
as  to  the  number  of  turns  per  inch  to  be  placed  in  yarn  varies 
in  different  mills,  but  is  always  based  on  multiplying  the 
square  root  of  the  number  being  spun  by  a  constant  number. 

In  American  mills,  for  ordinary  warp  yarns  this  constant 
number  is  4.75;  for  filling  yarn,  it  is  3.25. 

33.  To  find  the  twist  to  be  inserted  in  any  counts  of 
yarn: 

Rule. — Extract  the  square  root  of  the  coiuits  and  multiply 
the  result  by  the  standard  adopted. 

Example. — Find  the  turns  per  inch  in  number  25s  ordinary  warp. 

Solution. —  V25  =  5;  5  X  4.75  =  23.75,  turns  per  in.  Ans. 

It  should  be  understood  that  the  constants  given  above 
are  not  necessarily  adopted  in  all  cases,  since  varying 


§9 


YARN  CALCULATIONS,  COTTON 


t9 


circumstances  will  require  different  amounts  of  twist  in  the 
yarn.  These  constants,  however,  form  a  basis  on  which  to 
determine  the  amount  of  twist  that  is  to  be  placed  in  any 
counts  being  spun.  _ 


BREAKING  WEIGHT  OF  COTTON  WARP  YARN 


34.  An  instrument  that  is  commonly  adopted  for  finding 
the  strength  of  warp  yarns  is  shown  in  Fig.  3.  In  test¬ 
ing  the  strength  of  the  yarn  it 
is  the  custom  to  wrap,  or  reel, 

120  yards  and  place  this  yarn 
in  the  form  of  a  skein  on  the 
hooks  a,b,  Fig.  3.  By  turn¬ 
ing  the  handle  until  the  yarn 
breaks,  the  breaking  weight 
is  shown  by  means  of  the 
finger  and  dial.  When  finding 
the  breaking  weight  of  any 
counts  of  warp  yarn,  the  fol¬ 
lowing  rule  will  be  found  to  be 
of  advantage. 

35.  To  find  the  standard 
breaking  weight  of  warp  yarns: 

Rule. — Multiply  the  known 
coiaits  of  the  yarn  by  5.  Add  the 
constant  1,600.  Divide  the  sum 
by  the  counts  of  the  yarn.  The 
answer  is  the  approximate  stand¬ 
ard  breaking  weight,  in  pounds. 

Example. — What  is  the  standard 
breaking  weight  of  40s  warp  yarn? 

40  X  5  +  1,600 


Fig.  3 


Solution. — 


40 


=  45  lb.  Ans. 


The  same  number  of  yarn  from  the  same  mill  or  from 
the  same  mixing  of  cotton  will  not  show  an  absolutely 
uniform  breaking  strength.  This  being  the  case  it  can 
readily  be  understood  that  yarns  from  different  mills  or 


20 


YARN  CALCULATIONS,  COTTON 


§9 


from  different  mixings  of  cotton  will  always  show  a  vari¬ 
ation,  which  should,  however,  be  reduced  as  much  as 
possible. 

TABLE  IV 

BREAKING  WEIGHT  OF  AMERICAN  WARP  YARNS,  PER 
SKEIN.  WEIGHT  GIVEN  IN  POUNDS  AND  TENTHS 


Num- 

Breaking 

Num- 

Breaking 

Num- 

Breaking 

Num- 

Breaking 

ber 

Weight 

ber 

Weight 

ber 

Weight 

ber 

Weight 

I 

26 

66.3 

51 

36.6 

76 

25.8 

2 

27 

63.6 

52 

36.1 

77 

25-5 

3 

530.0 

28 

61.3 

53 

35-5 

78 

25-3 

4 

410.0 

29 

59-2 

54 

34-9 

79 

24.9 

5 

330.0 

30 

57-3 

55 

34-4 

80 

24 .6 

6 

275.0 

31 

55-6 

56 

33-8 

81 

24 -3 

7 

237.6 

32 

54-o 

57 

33-4 

82 

24.0 

8 

209.0 

33 

52.6 

58 

32.8 

83 

23-7 

9 

186.5 

34 

51.2 

59 

32-3 

84 

23-4 

10 

168.7 

35 

50.0 

60 

3i-7 

85 

23.2 

11 

1 54- 1 

36 

48.7 

61 

3i-3 

86 

22.8 

12 

142.0 

37 

47.6 

62 

30.8 

87 

22.6 

13 

I3I-5 

38 

46.5 

63 

30.4 

88 

22.4 

14 

122.8 

39 

45-5 

64 

30.0 

89 

22.2 

15 

1 15- 1 

40 

44.6 

65 

29.6 

90 

22.0 

16 

108.4 

4i 

43-8 

66 

29.2 

91 

2 1.7 

1 7 

102.5 

42 

43-o 

6  7 

28.8 

92 

21.5 

18 

97-3 

43 

42.2 

68 

28.5 

93 

21-3 

i9 

92.6 

44 

41.4 

69 

28.2 

94 

21.2 

20 

88.3 

45 

40.7 

70 

27.8 

95 

2 1 .0 

2  I 

83.8 

46 

40.0 

7 1 

27.4 

96 

20.7 

22 

79-7 

47 

39-3 

72 

27.1 

97 

20.5 

23 

75-9 

48 

38.6 

73 

26.8 

98 

20.4 

24 

72.4 

49 

37-9 

74 

26.5 

99 

20.2 

25 

69.2 

50 

37-3 

75 

26.2 

100 

20.0 

Table  IV  gives  what  is  known  as  Draper’s  standard 
for  the  breaking  weight  of  American  warp  yarns  and  is 


9 


YARN  CALCULATIONS,  COTTON 


21 


generally  used  in  the  United  States  as  a  standard,  being  based 
on  breaking  weights  of  samples  of  yarn  taken  from  a  large 
number  of  mills  and  over  extended  periods.  This  table 
may  be  used  for  reference,  although  the  rule  given  should 
be  memorized  for  use  when  it  is  not  convenient  to  refer 
to  the  table. 


METRIC  SYSTEM  OF  NUMBERING 

YARNS 

36.  In  recent  years  there  has  been  from  time  to  time 
considerable  agitation  along  the  lines  of  adopting  one  sys¬ 
tem  of  numbering  all  classes  of  yarns  used  in  textile  manu¬ 
facturing.  The  objects  of  these  movements  are:  first,  to 
bring  about  the  unification  of  the  method  of  stating  the 
degrees  of  fineness  of  the  yarns  from  all  varieties  of  fibers 
used  in  the  textile  industry  in  the  whole  world;  second,  to 
adopt  a  method  that  would  be  practical  in  every  way. 
The  advantages  of  such  a  system  as  this  are  apparent. 
The  chief  objection  to  it  is  that,  from  long  usage,  the 
methods  at  present  adopted  are  too  well  developed  for  a 
single  corporation  or  a  single  country  to  take  on  itself  such 
a  reform,  without  being  assured  that  its  neighbors  and 
competitors  would  simultaneously  do  the  same  thing,  aL 
a  mill  instituting  such  a  reform  would  be  at  a  disadvantage 
in  the  markets. 

The  method  usually  set  forth  is  that  of  numbering  all 
classes  of  yarns  by  what  is  known  as  the  metric  system, 
and  consists  of  considering  1  meter  of  No.  1  yarn  to  weigh  1 
gram,  the  meter  being  the  unit  of  length  in  the  metric  sys¬ 
tem  and  the  gram  being  the  unit  of  weight.  The  only  parts 
of  the  metric  system  that  it  is  necessary  to  understand  in 
order  to  convert  one  system  into  the  other,  are  the  equiva¬ 
lents  of  the  meter  and  the  gram,  which  are  as  follows: 

1  yard  =  .914  meter;  1  pound  =  453.59  grams 

37.  To  find  the  number  of  yarn  in  any  present  standard 
system  that  corresponds  to  the  number  of  yarn  in  the  metric 
standard  system: 


22 


YARN  CALCULATIONS,  COTTON 


9 


Rule. — Multiply  the  counts ,  given  in  the  metric  system ,  by 
453.59  ( grams  in  1  pound )  and  divide  by  the  standard  number 
of  yards  to  the  pound  in  the  present  system  multiplied  by  .914 
( meter  in  1  yard) . 

Example. — A  cotton  yarn  numbered  according  to  the  metric  system 
is  marked  40s.  Find  the  counts  in  the  present  system. 

0  40  X  453.59  00  ao. 

Solution.—  n,„  s „  —  23.631s.  Ans. 

840  X  .914 


38.  To  find  the  number  of  yarn  in  the  metric  standard 
system  that  corresponds  to  the  number  of  yarn  in  any 
present  standard  system: 

Rule. — Miiltiply  the  counts ,  given  in  the  present  system ,  by 
the  preseiit  standard  number  of  yards  to  the  pound  and  by  .914 
{meter  in  1  yard)  and  divide  by  453.59  {grams  in  1  pound) . 


Example. — A  cotton  yarn  numbered  according  to  the  present  sys¬ 
tem  is  marked  46s.  Find  the  counts  in  the  metric  system. 


46  X  840  X  .914 


77.86s.  Ans. 


Solution.— 


453.59 


YARN  CALCULATIONS, 
WOOLEN  AND  WORSTED 


WORSTED  YARNS 


SYSTEM  OF  NUMBERING  SINGEE  YARNS 

1.  From  the  earliest  times  of  the  manufacture  of  yarns 
it  has  been  found  necessary  to  adopt  some  system  by  means 
of  which  the  different  sizes  of  yarns  may  be  designated, 
since,  otherwise,  the  manufacturer  would  have  no  means  of 
readily  distinguishing  one  yarn  from  another.  The  sizes 
of  different  yarns  are  known  as  the  numbers,  or  counts,  of 
the  yarns.  The  words  number  and  counis  in  all  cases  repre¬ 
sent  a  certain  length  for  a  certain  weight. 

In  the  worsted  system  of  numbering  yarns,  all  counts  are 
based  on  the  standard  of  560  yards,  which  is  known  as 
1  hank.  In  other  words,  a  hank  of  worsted  yarn  contains 
560  yards.  That  this  is  a  standard  number  and  never  varies 
is  a  fact  that  should  always  be  remembered.  For  example, 
1  hank  of  a  certain  number  of  worsted  yarn  contains  560 
yards  and  a  hank  of  finer  or  coarser  yarn  also  contains  the 
same  number  of  yards. 

The  method  of  numbering  the  yarns  is  that  of  calling  a  yarn 
that  contains  1  hank,  or  560  yards,  in  1  pound  a  No.  1  yarn. 
If  the  yarn  contains  2  hanks,  or  1,120  yards,  in  1  pound,  it  is 
known  as  a  No.  2  yarn;  if  it  contains  3  hanks,  or  1,680  yards, 
in  1  pound,  it  is  known  as  a  No.  3  yarn.  Thus  it  will  be  seen 

For  notice  of  copyright,  see  page  immediately  following  the  title  page 


2  YARN  CALCULATIONS  §10 

that  the  number  of  hanks  (560  yards)  that  it  takes  to  weigh 
1  pound  determines  the  counts  of  the  yarn. 

The  counts  of  a  yarn  are  generally  indicated  by  placing 
the  letter  s  after  the  figure  representing  the  number  of  the 
yarn.  Thus,  26s  shows  the  counts  of  a  No.  26  yarn  and 
indicates  that  the  yarn  contains  26  hanks  (26  X  560  yards) 
in  1  pound. 


CALCULATIONS  OF  SINGLE  YARNS 

2.  There  are  several  calculations  that  must  be  considered 
when  dealing  with  single  worsted  yarns,  but  all  are  based  on 
one  equation,  namely, 

_ length  of  yarn,  in  yards _  _  ^ 

weight,  in  pounds  X  counts  X  standard 

The  standard  will  always  be  the  same  when  dealing  with 
worsted  (560  yards),  and  any  one  of  the  other  items  can  be 
found  by  simply  substituting  the  values  of  the  rest  in  the 
above  equation. 

3.  To  find  the  counts  of  a  yarn  when  the  length  and 
weight  are  given: 

Rule. — Divide  the  total  length  of  yarn ,  expressed  in  yards , 
by  the  product  of  the  weight ,  expressed  in  pounds, and  the  standard 
length . 

Example. — If  168,000  yards  of  yarn  weighs  10  pounds,  what  are 
the  counts? 


Solution. — 

168,000  (length  of  yarn,  in  yards) 

10  (weight,  in  pounds)  X  560  (standard) 


=  30s,  counts. 


Ans. 


4.  To  find  the  weight  of  yarn  when  the  length  and  counts 
are  known: 


Rule. — Divide  the  length ,  in  yards ,  by  the  product  of  the 
counts  a?id  the  standard  length. 

Example. — What  is  the  weight  of  42,000  yards  of  number  5s  yarn? 


42,000  (length,  in  yards) 

5  (counts)  X  560  (standard) 


15  lb.  Ans. 


Solution. — 


§10 


WOOLEN  AND  WORSTED 


3 


5.  To  find  the  length  of  yarn  when  the  weight  and  counts 
are  given: 

Rule. — Multiply  the  weight ,  in  pounds,  the  counts,  and  the 
standard  length  together. 

Example. — What  is  the  length  of  yarn  contained  in  a  bundle  that 
weighs  8  pounds,  the  counts  of  the  yarn  being  26s? 

Solution. —  8  (weight,  in  pounds)  X  26  (counts)  X  560  (standard) 
=  116,480  yd.  Ans. 

6.  When  numbering  yarns  several  different  weights  and 
also  names  for  lengths  will  be  found  to  be  used,  for  which 
reason  it  will  be  well  for  the  student  to  memorize  the  two 
following  tables.  These  tables  include  only  those  terms 
used  in  yarn  numbering. 


TABLE  I 

The  table  of  lengths  is  as  follows: 

1  yard  =  1  thread  (circumference  of  reel). 

80  yards  =  80  threads  =  1  lea. 

560  yards  =  560  threads  =  7  leas  =  1  hank. 

TABLE  II 

The  table  of  weights  is  as  follows: 

27.34  grains  =  1  dram. 

437.5  grains  =  16  drams  =  1  ounce. 

7,000.  grains  =  256  drams  =  16  ounces  =  1  pound. 

The  part  of  this  table  that  is  most  frequently  used  and 
which,  therefore,  should  be  carefully  noted  is  that  which 
states  that  there  are  7,000  grains  in  1  pound. 


SIZING  WORSTED  YARNS 

7.  It  is  generally  the  custom  in  mills  to  weigh  a  certain 
quantity  of  roving  or  yarn  from  each  machine  at  least  once 
a  day  and  by  this  means  ascertain  whether  it  is  being  kept 
at  the  required  weight.  This  process  is  known  as  sizing 
and  is  a  matter  that  should  always  be  very  carefully 
attended  to. 

From  the  rules  and  explanations  previously  given  it  will 
be  plain  that  if  560  yards,  or  1  hank,  were  always  the  length 


4 


YARN  CALCULATIONS, 


§10 


weighed,  in  order  to  learn  the  counts  of  the  yarn  it  would 
simply  be  necessary  to  divide  the  weight,  expressed  in 
pounds,  into  1  pound,  or  if  expressed  in  grains,  into  7,000 
(the  number  of  grains  in  1  pound).  It  will  readily  be  seen 
that  to  measure  off  560  yards  of  yarn  would  not  only  require 
considerable  time  but  would  also  produce  an  unnecessary 
waste  of  yarn.  To  overcome  these  difficulties,  when  sizing 
yarn  it  is  the  custom  to  measure  off  one-seventh  of  560 
yards,  or  80  yards;  to  weigh  this  amount;  and  divide  its 
weight,  in  grains,  into  one-seventh  of  7,000,  or  1,000. 


The  result  obtained  in  this  manner  will  be  the  same  as 
though  560  yards  were  taken  and  its  weight,  in  grains, 
divided  into  7,000. 

8.  The  Wrap  Reel. — When  sizing  yarns,  an  instrument 
known  as  a  wrap  reel  is  used  to  measure  the  yarn.  As  its 
name  indicates,  this  instrument  consists  of  a  reel,  generally 
1  yard  in  circumference.  The  yarn  is  wound  on  this  reel 
and  a  finger  indicates  on  a  disk  the  number  of  yards  reeled. 
Fig.  1  shows  an  ordinary  type  of  wrap  reel,  while  Fig.  2 
shows  yarn  and  roving  scales.  These  scales  are  suitable  for 
weighing  by  tenths  of  grains. 


§10 


WOOLEN  AND  WORSTED 


5 


Example. —  80  yards  of  yarn  is  reeled  and  found  to  weigh 
20  grains;  what  are  the  counts? 

Solution.—  1,000  4-  20  =  50s.  Ans. 

9.  From  the  foregoing  the  student  will  readily  under¬ 
stand  that  the  counts  of  a  yarn  may  be  obtained  by  finding 
the  weight  of  any  length  of  yarn  and  dividing  the  weight  into 
a  certain  number  of  grains,  this  number  having  the  same 
proportion  to  7,000  that  the  length  of  yarn-weighed  has  to  560. 


Fig.  2 


As  an  aid  to  the  student  in  any  calculations  that  he  may  have 
to  deal  with,  the  following  lengths  and  dividends  are  given: 


TABLE  III 


If  320  yards 
If  160  yards 
If  80  yards 
If  60  yards 
If  40  yards 
If  30  yards 
If  20  yards 
If  15  yards 
If  10  yards 
If  8  yards 
If  6  yards 
If  4  yards 
If  3  yards 
If  2  yards 
If  1  yard 


is  weighed, 
is  weighed, 
is  weighed, 
is  weighed, 
is  weighed, 
is  weighed, 
is  weighed, 
is  weighed, 
is  weighed, 
is  weighed, 
is  weighed, 
is  weighed, 
is  weighed, 
is  weighed, 
is  weighed, 


divide  weight, 
divide  weight, 
divide  weight, 
divide  weight, 
divide  weight, 
divide  weight, 
divide  weight, 
divide  weight, 
divide  weight, 
divide  weight, 
divide  weight, 
divide  weight, 
divide  weight, 
divide  weight, 
divide  weight, 


in  grains, 
in  grains, 
in  grains, 
in  grains, 
in  grains, 
in  grains, 
in  grains, 
in  grains, 
in  grains, 
in  grains, 
in  grains, 
in  grains, 
in  grains, 
in  grains, 
in  grains, 


into  4,000. 

into  2,000. 

into  1,000. 

into 

750. 

into 

500. 

into 

375. 

into 

250. 

into 

187*. 

into 

125. 

into 

100. 

into 

75. 

into 

50. 

into 

CO 

into 

25. 

into 

12*. 

6  YARN  CALCULATIONS,  §10 

SIZING  WORSTED  ROVING 

10.  Roving  is  a  term  used  to  designate  a  strand  of 
loosely  twisted  fibers  from  which  a  yarn  may  be  spun,  while 
the  term  yarn  is  applied  to  a  thread  composed  of  fibers  uni¬ 
formly  disposed  throughout  its  structure  and  having  a  certain 
amount  of  twist  for  the  purpose  of  enhancing  its  strength. 

It  is  usually  customary  to  designate  the  size  of  worsted 
roving  by  the  number  of  drams  that  40  yards  weighs;  thus, 
if  40  yards  of  a  certain  roving  weighs  3.2  drams,  it  is  known 
as  a  3.2-dram  roving.  When  very  fine  yarns  are  being  made, 
however,  the  roving  is  generally  designated  by  the  weight, 
in  drams,  of  80  yards,  this  length  being  taken  because  of  the 
greater  accuracy  obtained  by  weighing  a  greater  length.  In 
sizing  the  roving,  a  scale  capable  of  weighing  drams  and 
tenths  of  drams  is  commonly  used,  but  should  it  be  neces¬ 
sary  to  use  a  grain  scale,  the  measurements  given  in  Table  II 
would  prove  convenient. 

Since  worsted  yarn  is  numbered  on  the  basis  of  the  number 
of  hanks  of  560  yards  in  1  pound  and  roving  by  the  number 
of  drams  that  40  or  80  yards  weighs,  it  is  sometimes  desir¬ 
able  to  find  the  counts  of  a  roving. 

11.  To  find  the  counts  of  a  roving  when  its  weight,  in 
drams,  is  known: 

Rule. — If  the  weight ,  in  drams ,  of  40  yards  is  known,  divide 
this  weight  into  IS. 3.  If  the  weight ,  in  drams,  of  80  yards  is 
known,  divide  this  weight  into  36.6. 

Example. — What  are  the  counts  of  worsted  roving,  40  yards  of 
which  weighs  6.1  drams? 

Solution.—  18.3  4-  6.1  =  3s.  Ans. 

Note.—  18.3  and  36.6  are  known  as  jrauge  points  and  are  obtained  by  multi¬ 
plying  the  drams  in  1  pound  by  the  number  of  yards  weighed  (40  or  80)  and  dividing 
by  the  number  of  yards  in  1  hank,  thus: 

266  V  40 

- =  18.285.  practically  18.3 
560 

256  V  80 

=  36.57,  practically  36.6 
560 

From  this  it  will  be  seen  that  40  vards  of  number  Is  weighs  18.3  drams  and 
80  yards  of  number  Is  weighs  36.6  drams,  which  is  the  true  significance  of  the 
gauge  points. 


10 


WOOLEN  AND  WORSTED 


7 


EXAMPLES  FOR  PRACTICE 

1.  16,800  yards  of  worsted  yarn  weighs  1  pound;  what  are  the 

counts?  Ans.  30s 

2.  33,600  yards  of  worsted  yarn  weighs  5  pounds;  what  are  the 

counts?  Ans.  12s 


3.  80  skeins  of  worsted  yarn,  each  containing  2,100  yards,  weighs 

20  pounds;  what  are  the  counts?  Ans.  15s 

4.  What  are  the  counts  of  worsted  yarns  80  yards  of  which  weighs, 


respectively,  17,  21,  26,  and  32  grains? 


Ans.  ■ 


58.82s 

47.61s 

38.46s 

31.25s 


5.  If  a  cop  is  known  to  contain  960  yards  of  worsted  yarn 
and  35  of  these  cops  weigh  1  pound,  what  are  the  counts?  Ans.  60s 

6.  What  is  the  weight  of  37,000  yards  of  Is  worsted?  Ans.  66.07  lb. 

7.  What  is  the  weight,  in  ounces,  of  10,000  yards  of  36s  worsted? 

Ans.  7.93  oz. 

8.  How  many  yards  of  40s  worsted  are  there  in  24  pounds? 

Ans.  537,600  yd. 

9.  What  is  the  length  of  yarn  in  25  pounds  of  35s  worsted? 

Ans.  490,000  yd. 

10.  What  is  the  weight  of  168,000  yards  of  20s  worsted? 

Ans.  15  lb. 


PLY  YARNS 


METHOD  OF  NUMBERING 

12.  Definition. — Very  frequently  during  the  process  of 
manufacturing  yarns,  it  becomes  necessary  to  twist  two  or 
more  threads  together  to  form  one  coarser  thread.  Such 
yarns  are  commonly  known  as  ply  yarns,  also  as  folded,  or 
twisted,  yarns,  and  sometimes  as  double  and  twist  yarns. 
It  should  be  understood  that  when  yarns  are  folded  in  this 
manner  it  does  not  change  their  length,  with  the  exception 
of  a  slight  percentage  of  contraction,  which  takes  place 
during  the  twisting,  but  it  does  change  their  weight. 

The  method  of  numbering  worsted  ply  yarns  is  that  of 
giving  the  counts  of  the  single  yarns  that  are  folded  and 
placing  before  these  counts  the  number  that  indicates  the 


8 


YARN  CALCULATIONS, 


§10 


number  of  threads  folded.  Thus,  2  '40s  indicates  that  two 
threads  of  40s  single  yarn  are  folded  together,  causing  the 
ply  yarn  to  be  equal  in  weight  to  a  single  20s.  As  previ¬ 
ously  stated,  a  slight  contraction  takes  place  during  the 
process  of  twisting;  consequently,  to  make  the  resultant 
counts  20s,  the  single  yarns  folded  must  be  slightly  finer  than 
40s.  This  contraction,  however,  will  not  be  considered  in  the 
rules  and  examples  that  follow,  since  it  is  so  slight  that  it  is 
more  a  matter  of  experience  than  one  of  mathematics. 


CALCULATIONS  OF  PLY  YARNS 


13.  Folded  Yarns  of  tlie  Same  Counts. — It  is  not 
often  the  custom  in  mills  to  fold  yarns  of  different  counts, 
since  single  yarns  of  the  same  number  make  the  best  double, 
or  ply,  yarns.  Consequently,  when  yarns  of  the  same  counts 
are  folded,  in  order  to  find  the  counts  of  the  resultant  ply 
yarn,  it  is  simply  necessary  to  divide  the  counts  of  the 
yarns  folded  by  the  number  of  threads  that  constitute  the 
ply  yarn.  Thus,  for  example,  if  three  threads  of  a  45s  are 
folded  to  form  a  ply  yarn,  the  resultant  yarn  will  be  equiv¬ 
alent  in  weight  to  a  single  15s  (45  -4-  3  =  15). 

The  method  of  indicating  the  counts  of  this  ply  yarn 
is  3/ 45s.  The  rules  for  finding  the  counts,  weight,  and 
length  of  ply  yarns  are  similar  to  those  given  for  finding  the 
same  particulars  of  single  yarns,  with  the  exception  that  the 
counts  of  the  ply  yarn  do  not  indicate  the  actual  counts  of 
the  yarn,  but  instead  indicate  the  counts  of  the  yarns  folded; 
consequently,  when  figuring  to  find  these  particulars,  the 
actual  counts  of  the  ply  yarn  must  be  taken. 

Example  1. — What  is  the  weight  of  642,000  yards  of  2-ply  40s? 


Solution. — A  2-ply  40s  is  equal  in  weight  to  a  single  20s. 


642,000 
20  X  560 


57.32  lb.  Ans. 


Example  2. — What  is  the  length  of  20  pounds  of  2-ply  60s? 

Solution. —  2-ply  60s  is  equal  in  weight  to  a  single  30s. 

20  X  30  X  560  =  336,000  yd.  Ans. 


§10  WOOLEN  AND  WORSTED  9 

Example  3. — What  are  the  counts  of  a  2-ply  yarn,  176;400  yards  of 
which  weighs  10  pounds? 

o  176,400 

Solution.—  -  ■’  -  =  31.5s.  Ans. 

10  X  5b0 

This  answer  is  the  counts  of  the  ply  yarn  considered  as  a  single 
thread,  but  since  it  is  made  from  two  threads  of  the  same  counts  it 
will  be  a  2-ply  63s.  Therefore,  176,400  yd.  of  the  2-ply  63s  will  weigh 
10  lb. 

14.  Folded  Yarns  of  Different  Counts. — As  was 
stated,  it  does  not  often  occur  that  different  counts  are  folded 
together  to  form  a  ply  yarn,  since  singles  of  the  same  num¬ 
ber  make  the  best  double,  or  ply,  yarns.  However,  this  will 
sometimes  occur,  especially  when  it  is  desired  to  make  a 
fancy  yarn,  in  which  case  two  yarns  of  different  counts  are 
often  folded. 

Suppose,  for  illustration,  that  it  is  desired  to  find  the  result¬ 
ant  counts  of  a  40s  worsted  folded  with  a  20s  worsted.  Take 
as  a  basis  560  yards  of  each  yarn;  then  560  yards  of  the  40s 
weighs  iV  pound,  and  560  yards  of  the  20s  weighs  2V  pound. 
Consequently,  after  these  yarns  are  folded,  there  will  be 
560  yards  of  a  ply  yarn  the  weight  of  which  is  to  +  2V  =  1 \ 
pound. 

The  example  now  resolves  itself  into  the  following:  What 
are  the  counts  of  a  yarn  560  yards  of  which  weighs  -40  pound? 
Since  length  divided  by  (weight  times  standard)  equals 
560 

counts,  then  -5 - =  13.33s,  counts  of  the  ply  yarn. 

it)  X  560 

This  example  has  been  worked  out  to  some  length  in 
order  to  enable  the  student  to  understand  thoroughly  the 
method  of  numbering  ply  yarns.  The  following,  however, 
is  a  somewhat  shorter  method. 

15.  To  find  the  resultant  counts  when  two  yarns  of  dif¬ 
ferent  numbers  are  folded: 

Rule. — Multiply  the  two  counts  together  and  divide  the  result 
thus  obtained  by  the  sum  of  the  counts. 

Example. — Same  as  in  Art.  14. 

40  v  90 

Solution. —  i/TroTi  =  13.33s.  Ans. 

40  +  20 


10 


YARN  CALCULATIONS, 


10 


16.  Ply  Yarns  Composed  of  More  Than  Two 
Threads. — In  many  cases  it  will  be  found  necessary  to  find 
the  counts  of  a  ply  yarn  that  is  made  from  more  than  two 
single  threads.  Under  such  circumstances  it  will  be  neces¬ 
sary  to  follow  a  somewhat  different  process.  For  example, 
suppose  that  three  single  threads  of  24s,  36s,  and  72s, 
respectively,  are  folded  to  form  a  ply  yarn  and  that  it  is 
required  to  ascertain  the  counts  of  the  resultant  yarn.  This 
may  be  found  by  following  the  rule  previously  given  and 
performing  two  operation's  as  follows: 

First  find  the  counts  of  the  yarn  that  will  result  from 


=  14.4s.  The 


folding  the  24s  with  the  36s.  Thus: 


example  now  resolves  itself  into  the  following:  What  are 
the  counts  of  a  ply  yarn  made  from  one  thread  of  14.4s 

and  one  of  72s?  .14-4  X  72  _  ^  Ans. 

14.4  +  72 

A  somewhat  shorter  method  than  this,  however,  may  be 
applied  to  3  or  more  ply  yarns  made  from  different  counts. 

Rule. — Take  the  highest  counts  and  divide  it  by  itself  and 
by  each  qf  the  other  counts.  Add  the  results  thus  obtained  and 
divide  this  result  into  the  highest  cojcnts. 

Example. — Same  as  above. 

Solution. —  72  h-  72  =  1 


72  -s-  36  =  2 
72  -j-  24  =  3 


6 

72  4-  6  =  12s.  Ans. 

17.  To  find  the  resultant  counts  when  more  than  one  end 
of  the  different  counts  are  folded: 

Rule. — Divide  the  highest  counts  by  itself  and  by  each  of  the 
other  counts.  Multiply  the  result  in  each  case  by  the  number  of 
ends  of  that  counts.  Add  the  results  thus  obtained  and  divide 
this  result  into  the  highest  counts. 

Example. —  4  ends  of  80s  and  3  ends  of  60s  are  folded  to  form  a 
ply  yarn.  What  are  the  resultant  counts? 


§10 


WOOLEN  AND  WORSTED 


11 


Solution.—  80  -s-  80  =  1;  1  X  4  =  4 

80  4-  60  =  H;  1£  X  3  =  4 

8 

80  -f-  8  =  10s,  resultant  counts.  Ans. 

Another  calculation  that  will  frequently  occur  when  dealing 
with  ply  yarns  is  that  of  finding  the  counts  of  a  yarn  that 
must  be  folded  with  another  to  produce  the  given  counts. 

18.  To  find  what  counts  must  be  folded  with  another  to 
produce  a  given  counts: 

Rule. — Multiply  the  two  counts  together  and  divide  by  their 
difference. 


Example. — What  counts  of  yarn  must  be  folded  with  a  50s  to 
produce  a  30s  ply  yarn? 


Solution. — 


50  X  30 
50  -  30 


=  75s. 


Ans. 


This  example  may  be  proved  by  taking  the  two  single  yarns 
50s  and  75s  and  finding  the  ply  yarn  resulting  from  folding 
them,  following  the  rule  previously  given  for  finding  the 
counts  of  ply  yarn  resulting  from  the  folding  of  single  yarns. 

Still  another  calculation  that  enters  into  ply  yarns  is  that 
of  finding  the  weight  of  single  yarns  to  produce  a  given 
weight  of  ply  yarn  when  twisting  together  two  threads  of 
different  counts. 


19.  When  twisting  together  two  or  more  threads  of  dif¬ 
ferent  counts,  to  find  the  weight  of  each  required  to  produce 
the  given  weight: 

Rule. — Find  the  counts  resulting  from  folding  the  two  or 
more  threads;  then  as  the  counts  of  one  thread  is  to  the  resultant 
counts  so  is  the  total  weight  to  the  weight  required  of  that  thread. 


Example. — It  is  desired  to  produce  100  pounds  of  a  ply  yarn  from 
an  80s  and  a  32s;  what  weight  of  each  single  thread  must  be  used? 

o!!  =  22.85s,  resultant  counts. 

oU  -f-  oZ 

32  :  22.85  =  100  :  at 
22.85  X  100 


Solution. — 


x  = 


32 


=  71.40  lb.  of  32s.  Ans. 


It  will  readily  be  seen  that  in  order  to  find  the  weight  of  the  other 
thread  it  is  only  necessary  to  subtract  the  weight  of  the  32s  from  the 


12 


YARN  CALCULATIONS, 


§10 


total  weight,  or  100  lb.;  but,  in  order  to  enable  the  student  to  fully 
understand  this  rule,  the  required  weight  of  the  80s  will  be  worked  out 
similarly  to  that  of  the  32s.  Thus: 


x  = 


Resultant  counts  =  22.85s 
80  :  22.85  =  100  :  x 
22.85  X  100 


80 


=  28.56  lb.  of  80s. 


Ans. 


Note.— In  the  previous  example  the  weight  of  the  80s  yarn  plus  the  weight  of  the 
32s  yarn  should  equal  the  weight  of  the  ply  yarn,  but  owing  to  the  use  of  decimals, 
examples  of  this  kind  seldom  give  exact  results.  Thus.  71.40  pounds  +  28.56  pounds 
=  99.96  pounds:  whereas  the  total  weight  should  be  100  pounds. 


EXAMPLES  FOR  PRACTICE 

1.  If  a  thread  of  40s  and  one  of  60s  are  twisted  together,  what  are 

the  counts  of  the  resultant  ply  yarn?  Ans.  24s 

2.  A  20s,  30s,  and  60s  are  twisted  together;  what  are  the  counts  of 

the  ply  yarn?  Ans.  10s 

3.  What  are  the  counts  of  a  thread  that  must  be  twisted  with  a  44s 

to  produce  a  ply  yarn  equal  to  a  20s?  Ans.  36.66s 

4.  What  weight  of  40s  must  be  twisted  with  a  20s  to  produce 

200  pounds  of  ply  yarn?  Ans.  66.65  lb. 

5.  What  weight  of  100s,  80s,  and  60s  would  be  used  in  producing 

500  pounds  of  ply  yarn?  r  127.65  lb.  of  100s 

Ans.<  159.56  lb.  of  80s 
(212.75  lb.  of  60s 

6.  What  counts  of  ply  yarn  will  result  if  18s,  24s,  and  36s  are 
twisted  together,  and  what  weight  of  each  will  there  be  in  240  pounds 

of  the  folded  yarn?  r  106f  lb.  of  18s 

Ans.  8s<  80  lb.  of  24s 
I  53^  lb.  of  36s 

7.  A  ply  yarn  is  made  from  1  end  each  of  10s,  12s,  and  15s;  what 

are  the  counts  of  the  ply  yarn?  Ans.  4s 

8.  A  3-ply  yarn  is  made  from  80s,  40s,  and  30s,  and  weighs 

100  pounds;  what  weight  does  it  contain  of  each  count  of  yarn  and 
what  are  the  counts  of  the  ply  yarn?  f  17.64  lb.  of  80s 

Ans.  14.117s {  35.29  lb.  of  40s 
147.05  lb.  of  30s 

9.  What  will  the  resultant  counts  be  if  two  ends  of  40s  and  one  of 

60s  are  twisted  to  make  a  ply  yarn?  Ans.  15s 

10.  What  counts  result  from  twisting  a  60s  with  a  36s?  Ans.  22.5s 


§10 


WOOLEN  AND  WORSTED 


13 


WOOLEN  YARNS 


SYSTEMS  OF  NUMBERING  SINGLE  YARNS 


RUN  SYSTEM 

20.  The  common  method  of  numbering  woolen  yarns  is 
by  what  is  known  as  the  run  system.  With  this  system 

« 

the  number  of  runs  that  weigh  1  pound  indicates  the  size  of 
the  yarn,  a  run  being  1,600  yards.  For  example,  if  a  woolen 
yarn  is  a  4-run  yarn,  it  will  contain  4  X  1,600  yards  to 
1  pound. 

It  will  be  seen  that  this  method  of  numbering  the  yarn  is 
very  similar  to  that  explained  in  connection  with  worsted 
yarn,  the  only  difference  being  that  1,600  yards  in  this  case 
is  taken  as  the  standard  instead  of  560,  which  is  the  standard 
for  worsted.  Consequently,  the  rules  given  for  finding  the 
length,  weight,  and  counts  of  worsted  yarns  will  be  found  to 
apply  equally  well  in  this  case  with  the  exception  that  1,600  will 
be  used  in  place  of  560.  However,  in  order  that  the  student 
may  thoroughly  understand  the  application  of  these  rules, 
an  example  of  each  will  be  given. 

Example  1. — A  woolen  yarn  72,000  yards  in  length  is  found  to  weigh 
10  pounds;  what  run  is  it? 

.Solution.-  Io  VX600  =  4i'ru“-  Ans' 

Example  2. — What  is  the  weight  of  64,000  yards  of  2-run  woolen 
yarn? 

e  64,000  on  . 

SOLUTION.-  210^00  =  20  ,b'  Ans- 

Example  3. — What  is  the  length  of  5  pounds  of  a  6-run  yarn? 

Solution. —  5  X  6  X  1,600  =  48,000  yd.  Ans. 


14 


YARN  CALCULATIONS, 


§10 


21.  A  very  convenient  method  of  finding  the  weight, 
run,  or  length  of  a  woolen  yarn  when  the  weight  is 
expressed  in  ounces  may  be  explained  as  follows: 


Example  1. — What  is  the  weight,  in  ounces, 


51-run  yarn? 
Solution. — 


8,800 

5.5  X  1,600 


1  lb. 


of  8,800  yards  of  a 


But  it  will  be  noticed  that  it  is  desired  to  obtain  the  weight 
in  ounces;  therefore,  the  result  in  pounds  must  be  multiplied 
by  16,  which  gives  16  ounces  as  the  answer.  If,  instead  of 
multiplying  the  answer  by  16,  a  new  standard  is  obtained  by 
dividing  1,600  by  16,  the  result  will  be  the  same  but  the 
process  will  be  considerably  shortened.  Therefore,  when¬ 
ever  the  weight  is  expressed  in  ounces  the  standard 
employed  should  be  100. 


Example  2. — What  is  the  weight,  in  ounces,  of  5,250  yards  of  a 
3-run  yarn? 

5,250  1(7l 

Solution. —  ^  ^  =  17|  oz.  Ans. 


Example  3. — 
run  is  it? 

Solution. — 


4,000  yards  of  woolen  yarn  weighs  5  ounces. 


4,000 
5  X  100 


8-run.  Ans. 


What 


Example  4. — How  many  yards  are  there  in  10  ounces  of  a  3-run 
yarn  ? 

Solution. —  10  X  3  X  100  =  3,000  yd.  Ans. 


CUT  SYSTEM 

22.  In  the  vicinity  of  Philadelphia  the  counts  of  woolen 
yarns  are  based  on  what  is  termed  the  cut  system,  which 
has  for  its  standard  length  300  yards.  In  all  other  respects 
the  calculations  are  similar;  consequently,  the  rules  given 
will  apply  in  this  case  with  the  exception  that  300  is  used  as 
the  standard. 

Example  1. — What  is  the  size,  in  the  cut  system,  of  a  woolen  yarn 
36,000  yards  of  which  weighs  10  pounds? 

36,000 
10  X  300 


Solution. — 


=  12-cut.  Ans. 


WOOLEN  AND  WORSTED 


15 


8 10 


Example  2. — What  is  the  weight  of  66,000  yards  of  a  20-cut  yarn? 

c  66,000 

Solution.  20  X  300  =  U  lb'  Ans- 

Example  3. — What  is  the  length  of  10  pounds  of  a  24-cut  woolen 
yarn? 

Solution.—  10  X  24  X  300  =  72,000  yd.  Ans. 


SIZING  WOOLEN  ROVING  ANII  YARN 

23.  Both  woolen  roving  (in  some  mills  and  districts  this 
term  is  corrupted  into  roping)  and  yarn  are  sized  according 
to  the  number  of  runs  or  cuts  in  1  pound.  In  sizing  both 
the  yarn  and  roving  a  scale  known  as  a  run,  or  cut,  scale  is 


commonly  used.  This  scale,  shown  in  Fig.  3,  consists  sim¬ 
ply  of  a  brass  beam  that  is  carried  by  an  upright  standard 
and  so  arranged  that  when  a  given  number  of  yards  of 
roving  or  yarn  is  placed  on  the  beam  in  such  a  position  as 
to  balance  (the  beam  being  out  of  balance  when  not  in 
use),  the  position  of  the  yarn  will  indicate  its  size  on  a 
graduated  scale  on  the  beam.  In  order  that  the  reading 
may  be  as  accurate  as  possible,  the  yarn  or  roving  to  be 
weighed  is  twisted  into  a  small  bunch  and  suspended  by 
a  single  thread.  Run  or  cut  scales  are  usually  made  to 
indicate  the  correct  size  from  50  yards  of  yarn  or  roving, 
but  they  may  be  made  to  indicate  the  size  with  any  desired 
number  of  yards. 


16 


YARN  CALCULATIONS, 


§10 


If  other  than  the  number  of  yards  for  which  the  scale  is 
graduated  is  used  in  weighing,  the  reading  of  the  scale 
must  be  corrected.  In  order  to  do  accurate  work,  the 
equilibrium  of  a  run  or  cut  scale  should  be  adjusted  with 
a  known  weight  or  with  a  known  size  of  yarn.  This  can  be 
accomplished  by  turning  to  the  right  or  to  the  left  the  set¬ 
screw  in  the  heavy  end  of  the  beam. 

When  a  mill  is  not  equipped  with  a  run  scale,  the  size  of 
the  yarn  must  usually  be  determined  by  means  of  a  grain 
scale.  In  most  mills  20  yards  of  yarn  or  roving  is  measured 
off  for  sizing,  but  in  other  mills  50  yards,  100  yards,  or 
various  other  lengths  may  be  used. 

TABLE  IV 


Runs 

Grains 

Runs 

Grains 

1 

Runs 

Grains 

JL 

2 

175.00 

5 

17.50 

10 

8.75 

e 

8 

140.00 

5T 

16.66 

1O4 

8-53 

3 

4 

1 16.66 

52 

15.90 

1O2 

8-33 

7 

8 

100.00 

5f 

15.21 

107 

8.13 

i 

87.50 

6 

14.58 

1 1 

7-95 

14 

70.00 

67 

14.00 

I  I  4 

7.77 

1-2 

58.33 

62 

13.46 

1  ii 

7.60 

if 

50.00 

67 

12.96 

Ilf 

7-44 

2 

43.75 

7 

12.50 

12 

7.29 

27 

38.88 

-X 

7  4 

12.06 

124 

7.14 

27 

35-00 

72 

1 1 .66 

122 

7.00 

2! 

31.81 

7  f 

1 1.29 

I2f 

6.86 

3 

29.16 

8 

10.93 

13 

6.73 

3T 

26.92 

8| 

10.60 

I3T 

6.60 

32 

25.00 

8i 

10.29 

132 

6.48 

34 

23.33 

8f 

10.00 

1 3f 

6.36 

4 

21.87 

9 

9.72 

14 

6.25 

44 

20.58 

9i 

9-45 

142 

6.03 

42 

19.44 

92 

I  ^ 

9.21 

15 

5.83 

4f 

18.42 

3 

94 

8.97 

1 

§10 


WOOLEN  AND  WORSTED 


17 


24.  To  find  the  size  of  a  woolen  roving  or  yarn  when 
the  weight  of  a  given  number  of  yards  is  known: 

Rule. — Divide  the  number  of  yards  weighed ,  multiplied  by 
7,000 ,  by  the  standard  number  ( 1,600  or  300) ,  mtiltiplied  by  the 
weight,  in  grains. 

Example. — If  20  yards  of  woolen  yarn  weighs  25  grains,  what  is 
the  size  of  the  yarn? 

c  20  X  7,000  Q1 

Solution.—  ■  gn_  w  og  =  3i-run.  Ans. 

I,b00  X  25 

The  table  on  page  16  gives  the  size,  in  runs,  and  the  weight, 
in  grains,  of  20  yards  of  woolen  roving  or  yarn.  If  any  num¬ 
ber  of  yards  other  than  20  is  weighed,  the  reading  of  the 
table  will  have  to  be  corrected.  For  instance,  if  100  yards 
of  yarn  is  found  to  weigh  175  grains,  the  size  of  the  yarn  is 
not  l-run  but  five  times  (100  20  =  5)  i-run,  or  2i-run. 


PLY  YARNS 

25.  Ply  yarns  made  from  woolen  threads  are  num¬ 
bered  exactly  the  same  as  worsted  ply  yarns;  consequently, 
the  rules  given  in  connection  with  worsted  ply  yarns  will  be 
found  to  apply  equally  well  when  dealing  with  ply  yarns 
made  of  woolen  with  the  exception,  of  course,  that  in  each 
case  the  standard  number  of  yards  to  the  pound  used  in  the 
woolen  system  must  be  adopted. 


EXAMPLES  FOR  PRACTICE 

1.  Woolen  yarn  weighing  1  pound  contains  18,000  yards;  what 

is  the  number  in  both  run  and  cut  systems?  .  „  f  11.25-run 

Ans'  160-cut 

2.  What  is  the  weight,  in  ounces,  of  10,000  yards  of  woolen  yarn, 

the  yarn  being  4-run?  Ans.  25  oz. 

3.  What  is  the  length  of  75  pounds  of  2-run  woolen  yarn? 

Ans.  240,000  yd. 

4.  Find  the  weight  of  60,000  yards  of  4-cut  woolen  yarn. 

Ans.  50  lb. 


18 


YARN  CALCULATIONS, 


§  10 


5.  What  is  the  length  of  100  pounds  of  3-cut  woolen  yarn? 

Ans.  90,000  yd. 

6.  What  are  the  cut  counts  of  a  woolen  yarn  when  1,200  yards 

weighs  2  pounds?  Ans.  2-cut 

7.  What  are  the  run  counts  of  a  woolen  yarn  when  100,000  yards 

weighs  pounds?  Ans.  8.33-run 


BEAM  CALCULATIONS 

26.  The  yarn  that  forms  the  warp  of  a  fabric  after  being 
spun  is  wound  on  long  wooden  spools  known  as  jack,  or 
dresser,  spools.  The  yarn  from  a  number  of  these  spools 
is  then  passed  through  several  intermediate  processes  and 
ultimately  wound  on  the  loom  beam,  from  which  it  is 
slowly  unwound  in  the  loom  as  the  fabric  is  woven.  Whether 
the  yarn  is  on  the  jack-spool  or  the  loom  beam,  there  are 
several  important  rules  that  apply  equally  well  in  either  case. 

27.  To  find  the  size  of  yarn  on  a  jack-spool  or  beam  con¬ 
taining  only  one  size  of  yarn,  the  weight  of  the  yarn,  its 
length,  and  the  number  of  ends  being  known: 

Rule. — Multiply  the  length ,  in  yards,  of  the  warp  on  the 
beam  or  length  on  the  jack-spool  by  the  number  of  ends  on  the 
beam  or  spool ,  as  the  case  may  be.  Divide  the  result  thus 
obtained  by  the  weight  of  the  yarn,  in  pounds,  multiplied  by 
the  standard  number. 

Example  1. — A  worsted  warp  350  yards  long  contains  1,600  ends 
and  weighs  50  pounds;  what  is  the  number  of  the  yarn? 

0  350  X  1 ,600  on 

Solution. —  ..w  =  20s.  Ans. 

50  X  obO 

Example  2. — A  jack-spool  contains  200  yards  of  woolen  yarn  that 
weighs  2  pounds,  there  being  40  ends  on  the  spool;  what  is  the  size  of 
the  yarn? 

0  200  X  40  o1 

Solution.—  _  w  ,  =  2^-run.  Ans. 

2  X  1,600 

Note  — The  method  of  finding1  the  size  of  a  given  yarn  when  its  length  and  weight 
are  known  was  explained  in  Art.  3,  and  it  will  be  noticed  that  the  same  method  is 
adopted  in  the  above  rule,  the  only  difference  being  that  in  the  former  case  the  length 
of  only  one  thread  was  given,  while  in  this  case  there  are  a  number  of  threads,  or 
ends,  together,  thus  necessitating  the  multiplication  of  the  length  of  each  end  by  the 
number  of  ends  in  order  to  find  the  total  length  of  yarn.  The  rules  that  follow  are 
also  similar  to  those  employed  for  single  threads  except  that  a  number  of  ends  are 
dealt  with  instead  of  a  single  one. 


§10 


WOOLEN  AND  WORSTED 


19 


It  should  be  noted  in  connection  with  these  examples  that 
the  weight  of  the  spool  or  beam  has  not  been  taken  into  con¬ 
sideration,  the  weight  of  the  yarn  only  being  given.  In  the 
mill  the  yarn  and  the  spool,  or  beam,  on  which  it  is  wound 
are  usually,  of  necessity,  weighed  together;  in  which  case  it 
becomes  necessary  to  deduct  the  weight  of  the  beam  or 
spool  in  order  to  obtain  the  true  weight  of  the  yarn. 


Example  3. — A  woolen  warp  contains  1,200  ends  and  is  400  yards 
in  length;  the  warp  and  beam  together  weigh  140  pounds  and  the 
beam  alone  weighs  65  pounds.  What  is  the  size  of  the  yarn? 


Solution. — 


140  -  65  =  75  lb. 


400  X  1,200 
75  X  1,600 


4-ruu.  Ans. 


28.  To  find  the  number  of  ends  on  a  beam  when  the 
weight  and  length  of  the  warp  and  the  size  of  the  yarn  are 
known: 


Rule. — Multiply  the  weight ,  in  pounds ,  by  the  standard 
number  and  by  the  size  of  the  yarn.  Divide  the  result  thus 
obtained  by  the  length  of  the  warp ,  in  yards. 

Example  1. — A  worsted  warp  is  800  yards  long  and  weighs 
200  pounds  exclusive  of  the  beam.  If  the  warp  is  composed  of 
20s  yarn,  how  many  ends  does  it  contain? 


Solution. — 


200  X  560  X  20 
~  800 


2,800  ends.  Ans. 


Example  2.— A  woolen  warp  is  600  yards  long  and  weighs 
120  pounds  exclusive  of  the  beam.  If  the  warp  is  composed  of 
4^-run  yarn,  how  many  ends  does  it  contain? 


Solution. — 


120  X  1,600  X  4| 
600 


1,440  ends. 


29.  To  find  the  weight  of  a  warp  when  the  length,  num¬ 
ber  of  ends,  and  size  of  the  yarn  are  known: 

Rule. — Multiply  the  length  of  the  warp ,  in  yards ,  by  the 
number  of  ends  that  it  contains  and  divide  the  result  thus 
obtamed  by  the  standard  number  multiplied  by  the  size  of 
the  yarn. 


20 


YARN  CALCULATIONS, 


10 


Example  1. — A  woolen  warp  475  yards  long  contains  960  ends  of 
3-run  yarn;  what  is  the  weight  of  the  yarn? 


Solution.— 


475  X  960 
1,600  X  3 


95  lb.  Ans. 


Example  2. — A  worsted  warp  650  yards  long  contains  1,860  ends  of 
16s  yarn;  what  is  the  weight  of  the  yarn? 

650  X  1,860  00 

Solution. —  .  =  134.933  lb.  Ans. 

560  X  16 

30.  To  find  the  length  of  a  warp  when  the  weight,  num¬ 
ber  of  ends,  and  the  size  of  the  yarn  are  known; 


Rule. — Multiply  the  weight  of  the  warp ,  in  pounds ,  by  the 
standard  number  and  by  the  size  of  the  yarn ,  and  divide  the 
result  thus  obtained  by  the  member  of  ends  in  the  warp. 


Example  1. — A  worsted  warp  contains  2,400  ends  of  12s  yarn  and 
weighs  200  pounds;  how  long  is  it? 


Solution. — 


200  X  560  X  12 
2,400 


560  yd.  Ans. 


Example  2. — A  woolen  warp  contains  1,200  ends  of  2-run  yarn  and 
weighs  231  pounds;  how  long  is  it? 


Solution. — 


231  X  1,600  X  2 

1,200 


616  yd.  Ans. 


31.  To  find  the  length  of  warp  that  can  be  placed  on  a 
beam: 


Rule. — Find  the  weight  of  yam  that  the  beam  will  contain , 
by  weighing  a  beam  of  the  same  size  wheel  filled  with  yarn  and 
deducting  the  weight  of  the  beam  itself.  Then  apply  the  rule  in 
Art.  30. 


Example  1. — A  certain  size  beam  when  filled  with  yarn  weighs 
140  pounds,  the  beam  itself  weighing  50  pounds.  What  length  of  a 
warp  composed  of  1,800  ends  of  28s  worsted  can  be  placed  on  it? 


Solution. — 


140  —  50  =  90  lb.  of  yarn. 


90  X  560  X  28 
1,800 


=  784  yd.  Ans. 


Example  2. — Same  beam  as  in  Example  1.  How  many  yards  of  a 
warp  containing  3,000  ends  of  40s  worsted  can  be  placed  on  it? 

Solution. —  140  —  50  =  90  lb.  of  yarn. 

90  X  560  X  40 


§10 


WOOLEN  AND  WORSTED 


21 


EXAMPLES  FOR  PRACTICE 

1.  A  worsted  warp  and  beam  weigh  170  pounds,  the  beam  alone 

weighing  70  pounds.  If  the  warp  is  350  yards  long  and  contains 
2,240  ends,  what  is  the  size  of  the  yarn?  Ans.  14s. 

2.  A  woolen  warp  is  320  yards  long  and  weighs  124  pounds,  exclu¬ 

sive  of  the  beam.  If  the  warp  is  composed  of  2|-run  yarn,  how  many 
ends  does  it  contain?  Ans.  1,550  ends 

3.  A  worsted  warp  780  yards  long  contains  2,480  ends  of  20s  yarn; 

what  is  the  weight  of  the  yarn?  Ans.  172.714  lb. 

4.  A  woolen  warp  contains  824  ends  of  2i-run  yarn  and  weighs 

103  pounds.  How  long  is  the  warp?  Ans.  450  yd. 


FANCY  WARPS 

32.  It  frequently  happens  that  in  the  production  of  fancy 
goods  there  are  two  or  more  colors  of  yarn  used  in  a  single 
warp  in  order  to  obtain  a  certain  effect  in  the  fabric.  When 
this  is  the  case,  it  often  necessitates  the  finding  of  the  total 
number  of  ends  of  each  color  in  the  warp  or  the  weight  of 
each  particular  yarn,  etc.  When  the  yarn  is  on  the  loom 
beam  the  following  rule  will  be  found  of  value. 

33.  To  find  the  number  of  ends  of  each  color  of  yarn  on 
a  beam  when  the  pattern  of  the  warp  and  the  total  number  of 
ends  in  the  warp  are  known: 


Rule. — Multiply  the  number  of  ends  of  any  one  color  in  the 
pattern  by  the  total  number  of  ends  on  the  beam  and  divide  by 
the  total  number  of  ends  in  the  pattern. 


Example  1. — A  warp  is  arranged,  or  dressed,  16  ends  of  black, 
8  ends  of  slate,  20  ends  of  black,  and  4  ends  of  white,  and  contains 
2,400  ends;  how  many  ends  of  each  color  are  there? 


Solution. — In  this  example  there  are  16  -f-  8  -f  20  +  4  =  48  ends 
in  one  repeat  of  the  pattern,  of  which  36  ends  are  black;  8,  slate;  and 
4,  white. 


36  X  2,400 
48 

8  X  2,400 
48 

4  X  2,400 


1,800  ends  of  black 
400  ends  of  slate 
200  ends  of  white 


■Ans. 


48 


22 


YARN  CALCULATIONS 


Proof. —  1,800  +  400  +  200  =  2,400  ends  in  warp. 


Note. — One  repeat  of  the  colors  or  counts  in  a  warp  is  known  as  the 
pattern  of  the  warp.  For  an  illustration,  suppose  the  ends  on  a 
beam  are  arranged  8  black,  2  white,  8  black,  2  slate.  This  would  be 
the  pattern  of  the  warp  and  all  the  other  ends  arranged  in  the  same 
manner  would  simply  form  repeats  of  the  pattern. 

After  the  total  number  of  ends  of  each  color  are  found,  it 
is  only  necessary  to  apply  the  rule  in  Art.  29  in  order  to  find 
the  weight  of  each  color  in  the  warp,  the  length  of  the  warp 
and  the  size  of  the  yarn  being  known. 

Example  2. — If  the  warp  in  example  1  is  640  yards  long  and  is 
composed  of  20s  worsted  yarn,  how  many  pounds  of  each  color  of 
yarn  does  it  contain? 

Solution. — 


640  X  1,800 
560  X  20 
640  X  400 
560  X  20 
640  X  200 
560  X  20 


=  102.857  lb.  of  black  yarn 


=  22.857  lb.  of  slate  yarn  >Ans. 


=  11.428  lb.  of  white  yarn 


34.  When  warps  are  composed  of  yarns  of  different 
counts  arranged  in  a  pattern,  the  same  method  may  be 
employed  to  find  the  number  of  ends  of  each  count,  after 
which  the  rule  in  Art.  29  may  be  applied  to  find  the  weight 
of  yarn  of  each  count. 

Example. — A  worsted  warp  is  to  be  made  containing  3,000  ends 
and  432  yards  long.  The  yarn  used  is  of  two  counts  arranged  12  ends 
of  28s,  2  ends  of  6s,  4  ends  of  28s,  2  ends  of  6s.  How  many  pounds  of 
each  yarn  is  necessary  for  the  warp? 

Solution. — In  this  example  there  are  12  +  2  +  4  +  2  =  20  ends  in 
each  pattern,  of  which  16  are  of  28s  yarn  and  4  of  6s  yarn. 


16  X  3,000 
20 


=  2,400  ends  of  28s 


4  X  3,000 
20 


=  600  ends  of  6s 


432  X  2,400 
560  X  28 


432  X  600  ' 

-660W  =  77 


=  66.122  lb.  of  28s 


>Ans. 


=  77.142  lb.  of  6s 


§10 


WOOLEN  AND  WORSTED 


23 


AVERAGE  NUMBERS 

35.  When  a  warp  contains  yarn  of  more  than  one  size,  it 
is  often  of  advantage  to  find  the  average  size  of  the  warp  in 
order  that  other  calculations  may  be  more  easily  performed. 
By  the  term  average  number  of  the  warp,  which  is  often 
used  in  this  connection,  a  number  of  yarn  is  meant  that  will 
be  of  the  same  weight  as  the  yarns  of  unequal  counts  that  are 
used  in  the  warp,  if  the  same  number  of  ends  are  used. 

36.  To  find  the  average  size  of  warp  yarn  in  a  warp  that 
contains  two  or  more  sizes: 

Rule. — Divide  the  total  number  of  ends  of  each  counts  by  its 
size.  Add  the  results  thus  obtained  arid  divide  the  sum  into  the 
total  number  of  ends  in  the  warp. 

Example. — Suppose  that  on  the  same  beam  there  are  1,600  ends  of 
40s  worsted  and  800  ends  of  10s.  It  is  desired  to  know  what  number 
of  yarn  will  make  a  warp  of  the  same  weight  per  yard  with  the  same 
number  of  ends. 

Solution. —  1,600  4  40  =  40 

800  4  10  =  80 

2+>0  120 

2,400  4  120  =  20s,  average  number.  Ans. 

Proof. — That  the  above  example  is  correct  may  be  readily 
shown  by  finding  the  weight  of  1  yard  of  warp  composed  of 
1,600  ends  of  40s  and  800  ends  of  10s  worsted  and  also  the 

weight  of  1  yard  of  a  warp  composed  of  2,400  ends  of  20s 

worsted. 

=  .0714  lb.,  weight  of  40s  yarn  in  first  warp 

=  .1428  lb.,  weight  of  10s  yarn  in  first  warp 

=  .2142  lb.,  weight  of  1  yard  of  first  warp 
=  .2142  lb.,  weight  of  1  yard  of  second  warp 

From  this  it  will  be  seen  that  the  weight  of  the  original 
warp  and  that  of  the  one  composed  of  the  same  number  of 
ends  of  20s  is  identical;  therefore,  the  example  must  be  correct. 


1X1,600 
560  X  40 
1  X  800 
560  X  10 
.0714  +  .1428 
IX  2,400 
560  X  20 


24 


YARN  CALCULATIONS,  §10 

37.  The  rule  just  given  will  be  found  to  apply  equally 
well  if  more  than  two  sizes  of  warp  yarn  are  used. 

Example. — A  warp  is  composed  of  1,200  ends  of  40s  worsted, 
600  ends  of  20s,  and  120  ends  of  10s.  What  is  the  average  number? 

Solution. —  1,200  40  =  30 

600  -f-  20  =  30 

120  -p  10  =  12 

L920  72 

1,920  -f-  72  =  26.66s,  average  number.  Ans. 

38.  In  cases  where  the  order  of  arranging  the  different 
counts  of  yarn  in  the  warp  is  known  but  the  total  number  of 
ends  is  not  given,  the  same  rule  will  be  found  to  apply  if 
the  number  of  ends  in  one  pattern  of  the  warp  is  considered 
as  the  total  number  of  ends. 

Example. — A  warp  is  arranged  48  ends  of  36s  and  2  ends  of  10s. 
Find  the  average  number. 

Solution. —  48  -s-  36  =  1.333 

2  -5-  10  =  .2 

50  1 .533 

50  -r*  1.533  =  32.615s,  average  number.  Ans. 


EXAMPLES  FOR  PRACTICE 

1.  A  worsted  warp  contains  1,620  ends  of  45s  yarn  and  840  ends  of 

20s  yarn;  find  the  average  counts  of  the  warp.  Ans.  31.53s 

2.  A  woolen  warp  contains  900  ends  of  44-run  yarn  and  450  ends 

of  24-run;  find  the  average  size  of  the  yarn.  Ans.  3f-run 


§10 


WOOLEN  AND  WORSTED 


25 


METRIC  SYSTEM  OF  NUMBERING 

YARNS 

39.  In  recent  years  there  has  been  from  time  to  time 
considerable  agitation  along  the  lines  of  adopting  one 
system  of  numbering  all  classes  of  yarns  used  in  textile 
manufacturing.  The  objects  of  these  movements  are:  first, 
to  bring  about  the  unification  of  the  method  of  stating  the 
degree  of  fineness  of  the  yarns  for  all  varieties  of  fibers  used 
in  the  textile  industry  in  the  whole  world;  second,  to  adopt 
a  method  that  would  be  practical  in  every  way. 

The  advantages  of  such  a  system  as  this  would  be  many. 
The  chief  objection  to  it  is  that,  from  long  usage,  the  methods 
at  present  adopted  are  too  well  developed  for  a  single  corpo¬ 
ration  or  a  single  country  to  take  on  itself  such  a  reform, 
without  being  assured  that  its  neighbors  and  competitors 
would  simultaneously  do  the  same  thing,  as  a  mill  instituting 
such  a  method  would  be  at  a  disadvantage  as  compared  with 
its  competitors. 

The  method  usually  set  forth  is  that  of  numbering  all 
classes  of  yarns  by  what  is  known  as  the  metric  system 
and  consists  of  considering  1  meter  of  number  Is  yarn  to 
weigh  1  gram,  the  meter  being  the  unit  of  length  in  the 
metric  system  and  the  gram  being  the  unit  of  weight.  The 
only  parts  of  the  metric  system  that  it  is  necessary  to  under¬ 
stand  in  order  to  convert  one  system  into  the  other  are  the 
equivalents  of  the  meter  and  the  gram,  which  are  as  follows: 

1  yard  =  .914  meter,  1  pound  =  453.59  grams 

40.  To  find  the  number  of  yarn  in  any  present  standard 
system  that  corresponds  to  the  number  of  yarn  in  the  metric 
standard  system: 

Rule. — Multiply  the  counts,  given  in  the  metric  system ,  by 
453.59  ( grams  in  1  pound )  and  divide  by  the  standard  number 


26 


YARN  CALCULATIONS, 


§10 


of  yards  to  the  pound  in  the  present  system  multiplied  by  .914 
( meter  in  1  yard). 

Example. — A  worsted  yarn  numbered  according  to  the  metric 
system  is  marked  40s.  Find  the  counts  in  the  present  system. 


40  X  453.59 
560  X  .914 


=  35.447s.  Ans. 


Solution. — 


41.  To  find  the  number  of  yarn  in  the  metric  standard 
system  that  corresponds  to  the  number  of  yarn  in  any  pres¬ 
ent  standard  system: 

Rule. — Multiply  the  counts ,  given  in  the  present  system ,  by 
the  present  standard  number  of  yards  to  the  pound  and  by  .914 
( meter  in  1  yard)  and  divide  by  453.59  (grams  in  1  pound). 

Example. — A  worsted  yarn  numbered  according  to  the  present 
system  is  marked  46s.  Find  the  counts  in  the  metric  system. 


Solution. — 


46  X  560  X  .914 
453.59 


=  51.907s.  Ans. 


INTRODUCTION 

1.  From  the  earliest  times  of  the  manufacture  of  yarns  it 
has  been  found  necessary  to  adopt  some  system  by  means  of 
which  the  different  sizes  of  yarns  can  be  designated,  since 
otherwise  the  manufacturer  has  no  means  of  readily  distin¬ 
guishing  one  yarn  from  another.  As  no  uniform  standard 
has  been  adopted,  the  methods  used  in  the  various  manufac¬ 
turing  centers  of  the  world  have  become  too  numerous  to 
allow  a  detailed  description  being  given  here;  therefore, 
only  those  found  in  common  practice,  with  particular  refer¬ 
ence  to  the  American  methods,  will  be  dealt  with. 

The  different  yarns  employed  in  textile  manufactures  may 
be  divided  into  several  classes:  For  example,  they  may  be 
divided  into  (1)  vegetable  fibers ,  which  include  cotton,  linen, 
jute,  etc.;  (2)  animal  fibers ,  which  include  woolen,  worsted, 
silk,  etc.  Again,  these  yarns  may  be  divided  with  reference 
to  the  system  employed  when  numbering  them,  as,  for 
example,  one  class  may  be  said  to  be  that  which  includes  all 
yarns  based  on  the  system  of  having  higher  numbers  for 
coarser  threads;  in  this  class  are  included  coarse  linen,  coarse 
jute,  and  raw  silk.  The  second  class  includes  all  yarns  based 
on  the  system  of  having  higher  numbers  for  finer  threads. 
This  method  will  be  found  to  be  the  more  common  and 
includes  cotton,  woolen,  worsted,  mohair,  spun  silk,  fine 
linen,  and  fine  jute. 

For  notice  of  copyright,  see  page  immediately  following  the  title  page 

l  11 


2 


YARN  CALCULATIONS,  GENERAL 


§11 


2.  The  words  member  and  counts  when  applied  to  yarns 
represent  in  all  cases  a  certain  length  for  a  certain  weight. 
Other  terms  frequently  used  are  run,  cut ,  skein ,  spindle ,  grist , 
etc.  A  person  figuring  counts  in  any  mill,  especially  when 
dealing  with  cloth  calculations,  will  find  it  necessary  to 
understand  the  systems  employed  when  numbering  different 
yarns,  such  as  wool,  cotton,  worsted,  and  silk,  for  which 
reason  the  numbering  of  these  will  be  thoroughly  dealt  with, 
each  one  being  considered  by  itself. 


SINGLE  YARNS 


COTTON  YARNS 


NUMBERING  SYSTEM 

3.  Method  of  Determining  tlie  Counts.  —  In  the 

cotton  system  of  numbering  yarns  all  counts  are  based  on 
the  standard  of  840  yards,  which  is  known  as  1  hank;  in 
other  words,  a  hank  of  cotton  yarn  contains  840  yards,  with¬ 
out  regard  to  the  fineness  or  coarseness  of  the  yarn.  This 
is  a  constant  number,  and  the  fact  that  it  never  varies  should 
always  be  remembered. 

The  method  of  numbering  the  yarns  is  that  of  calling  a 
yarn  that  contains  1  hank,  or  840  yards,  in  1  pound  a  No.  1 
yarn.  If  the  yarn  contains  2  hanks,  or  1,680  yards,  in 
1  pound,  it  is  known  as  a  No.  2  yarn;  if  it  contains  3  hanks, 
or  2,520  yards,  in  1  pound,  it  is  known  as  a  No.  3  yarn. 
Thus  it  will  be  seen  that  the  number  of  hanks  (840  yards)  it 
takes  to  weigh  1  pound  determines  the  counts  of  the  yarn. 

The  counts  of  a  yarn  are  generally  indicated  by  placing  a 
letter  s  after  the  figure  representing  the  number  of  the  yarn; 
thus,  26s  shows  the  counts  of  a  yarn  and  indicates  that  the 
yarn  contains  26  hanks  (26  X  840  yards)  in  1  pound. 


11 


YARN  CALCULATIONS,  GENERAL 


3 


CALCULATIONS  OF  SINGLE  COTTON  YARNS 

4.  Several  calculations  become  necessary  when  dealing 
with  single  yarns,  but  all  are  based  on  one  equation;  namely, 

length  of  yarn,  in  yards  _  ^ 

weight,  in  pounds  X  counts  X  standard 

The  standard  will  always  be  the  same  when  dealing  with 
cotton  (840  yards),  and  any  one  of  the  other  items  can  be 
found  by  simply  substituting  the  values  of  the  rest  in  the 
above  equation. 


5.  To  find  the  counts  of  a  yarn  when  the  length  and 
weight  are  given: 

Rule. — Divide  the  total  length  of  yam,  expressed  in  yards, 
by  the  product  of  the  weight,  expressed  in  pounds,  and  the  stand¬ 
ard  length. 

Example. —  168,000  yards  of  yarn  is  found  to  weigh  5  pounds; 
what  are  the  counts? 


Solution. — 

168,000  (length  of  yarn,  in  yards)  _ 

5  (weight,  in  pounds)  X  840  (standard) 


=  40s  counts. 


Ans. 


6.  To  find  the  weight  of  yarn  when  the  length  and  counts 
are  known: 


Rule. — Divide  the  length,  in  yards,  by  the  product  of  the 
couiits  and  the  standard  length. 


Example. — What  is  the  weight  of  42,000  yards  of  number  5s  yarn? 


.Solution. — 


42,000  (length,  in  yards) 

5  (counts)  X  840  (standard) 


10  lb.  Ans. 


7.  To  find  the  length  of  yarn  when  the  weight  and  counts 
are  given: 

Rule. — Multiply  the  weight,  in  pounds ,  the  cowits,  and  the 
standard  length  together. 

Example. — What  length  of  yarn  is  contained  in  a  bundle  weigh¬ 
ing  8  pounds,  the  counts  of  the  yarn  being  26s? 

Solution. —  8  (weight,  in  pounds)  X  26  (counts)  X  840  (standard)  = 
174,720  yd.  Ans. 


4 


YARN  CALCULATIONS,  GENERAL 


§11 


8.  From  the  rules  and  examples  given,  the  student  will 
readily  see  that  there  is  one  primary  fact  that  should  con¬ 
stantly  be  kept  in  mind  when  trying  to  understand  the  system 
of  cotton-yarn  numbering-;  viz.,  in  all  cases  the  number  of 
hanks  (840  yards)  of  any  yarn  that  it  takes  to  weigh  1  pound 
is  the  counts  of  that  yarn. 

When  numbering  yarns,  it  will  be  found  that  several 
different  weights  and  names  of  lengths  are  used,  for  which 
reason  it  will  be  well  to  memorize  the  two  following  tables. 
These  tables  include  only  those  terms  used  in  yarn  numbering. 

TABLE  I 

The  table  of  lengths  is  as  follows: 

1|  yards  =  1  thread,  or  circumference  of  wrap  reel. 

120  yards  =  80  threads  =  1  skein,  or  lea. 

840  yards  =  500  threads  =  7  skeins,  or  leas  =  1  hank. 

TABLE  II 

The  table  of  weights  is  as  follows: 

27.34  grains  =  1  dram. 

437.5  grains  =  16  drams  =  1  ounce. 

7,000  grains  =  256  drams  =  16  ounces  =  1  pound. 

The  part  of  this  table  most  frequently  used  and  which, 
therefore,  should  be  carefully  noted  is  that  which  states 
that  there  are  7,000  grains  in  1  pound. 


SIZING  COTTON  ROVING  AND  YARN 

9.  Roving  is  a  term  used  to  designate  a  strand  of  loosely 
twisted  fibers  from  which  a  yarn  may  be  spun,  while  the  term 
yarn  is  applied  to  a  thread  composed  of  fibers  uniformly 
disposed  throughout  its  structure  and  having  a  certain 
amount  of  twist  for  the  purpose  of  enhancing  its  strength. 

Some  method  must  be  adopted  in  every  well-regulated 
mill  by  means  of  which  the  exact  counts  of  the  roving  and 
yarn  being  run  may  be  known.  It  is  generally  the  custom 
to  weigh  a  certain  quantity  from  each  machine  at  least  once 
a  day  and  by  this  means  ascertain  whether  the  roving  or 
yarn  is  being  kept  at  the  required  weight.  This  process  is 
known  as  sizing  and  is  a  matter  that  should  always  be  very 
carefully  attended  to. 


§11 


YARN  CALCULATIONS,  GENERAL 


5 


10.  The  Wrap  Reel. — When  sizing  yarns,  an  instru¬ 
ment  known  as  a  wrap  reel  is  used  to  measure  the  yarn. 


As  its  name  indicates,  this  instrument  consists  of  a  reel, 
generally  I2  yards  in  circumference.  The  yarn  is  wound 


on  this  reel  and  a  finger  indicates  on  a  disk  the  number 
of  yards  reeled.  Fig.  1  shows  an  ordinary  type  of  wrap 


6 


YARN  CALCULATIONS,  GENERAL 


§11 


reel,  while  Fig.  2  shows  yarn  and  roving  scales.  These 
scales  are  suitable  for  weighing  by  tenths  of  grains. 

In  case  the  standard  length  of  the  yarn  being  sized  was 
reeled  off,  all  that  would  be  necessary,  in  order  to  find  the 
counts,  would  be  to  divide  the  weight  into  1  pound.  This 
would,  however,  require  considerable  time,  in  addition  to 
causing  an  unnecessary  amount  of  waste  yarn.  It  is,  there¬ 
fore,  customary  to  reel  off  a  certain  number  of  yards  and 
divide  the  weight  of  this  amount,  expressed  in  grains,  into 
the  number  of  grains  that  bears  the  same  relation  to  7,000 
(grains  in  1  pound)  that  the  number  of  yards  weighed  bears 
to  the  standard  length.  For  example,  if  it  is  desired. to  size 
the  yarn  running  on  a  cotton  spinning  frame,  the  bobbin  of 
yarn  is  removed  from  the  frame  and  taken  to  the  wrap  reel. 
120  yards  is  reeled  off  and  the  weight,  in  grains,  of  this 
120  yards  divided  into  1,000,  the  answer  being  the  counts  of 
the  yarn.  This  must  be  true,  because  120  yards  bears  the 
same  relation  to  840  yards  (the  standard)  that  1,000  bears 
to  7,000  (the  number  of  grains  in  1  pound). 

Example. —  120  yards  of  cotton  yarn  is  reeled  and  found  to 

weigh  40  grains;  what  are  the  counts? 

Solution. —  1,000  -p  40  =  25s.  Ans. 

11.  This  basis  of  indicating  the  size  of  the  yarn  is  also 
used  to  designate  the  size  of  roving,  although  when  sizing 
roving,  a  shorter  length  is  used.  It  is  customary  in  this 
case  to  measure  off  one-seventieth  of  840  yards,  or  12  yards, 
and  divide  the  weight,  in  grains,  of  this  length  of  roving 
into  one-seventieth  of  7,000,  or  100. 

Example. —  12  yards  of  roving  is  found  to  weigh  20  grains;  what 

are  the  counts? 

Solution. —  100  -p  20  =  5-hank.  Ans. 

The  counts  of  roving  are  indicated  in  a  somewhat  different 
manner  from  the  counts  of  yarn.  Thus,  in  this  case,  the 
roving  would  be  known  as  5-hank  roving ,  indicating  that 
5  hanks  weigh  1  pound. 

12.  From  the  foregoing,  the  student  will  readily  under¬ 
stand  that  the  counts  of  a  cotton  yarn  may  be  obtained  by 


11 


YARN  CALCULATIONS,  GENERAL 


7 


finding  the  weight  of  any  length  of  yarn,  and  dividing  the 
weight  into  a  number  of  grains  that  has  the  same  propor¬ 
tion  to  7,000  that  the  length  of  yarn  weighed  has  to  840. 
As  an  aid  to  the  student  the  following  lengths  and  dividends 
are  given: 

TABLE  III 

If  480  yards  is  weighed,  divide  weight,  in  grains,  into  4,000. 

If  240  yards  is  weighed,  divide  weight,  in  grains,  into  2,000. 

If  120  yards  is  weighed,  divide  weight,  in  grains,  into  1,000. 

If  60  yards  is  weighed,  divide  weight,  in  grains,  into  500. 

If  40  yards  is  weighed,  divide  weight,  in  grains,  into  3334- 

If  30  yards  is  weighed,  divide  weight,  in  grains,  into  250. 

If  20  yards  is  weighed,  divide  weight,  in  grains,  into  166f. 

If  15  yards  is  weighed,  divide  weight,  in  grains,  into  125. 

If  12  yards  is  weighed,  divide  weight,  in  grains,  into  100. 

If  10  yards  is  weighed,  divide  weight,  in  grains,  into  834- 

If  8  yards  is  weighed,  divide  weight,  in  grains,  into  66f. 

If  6  yards  is  weighed,  divide  weight,  in  grains,  into  50. 

If  4  yards  is  weighed,  divide  weight,  in  grains,  into  334- 

If  3  yards  is  weighed,  divide  weight,  in  grains,  into  25. 

If  2  yards  is  weighed,  divide  weight,  in  grains,  into  16f. 

If  1  yard  is  weighed,  divide  weight,  in  grains,  into  84. 


WOOLEN  YARNS 


RUN  SYSTEM  OF  NUMBERING 

13.  The  common  method  of  numbering  woolen  yarns  is 
by  what  is  known  as  the  run  system.  With  this  system 
the  number  of  runs  that  weigh  1  pound  indicates  the  size  of 
the  yarn,  a  run  being  1,600  yards.  Thus,  for  example,  if  a 
woolen  yarn  is  a  4-run  yarn,  it  will  contain  4  X  1,600  yards 
to  1  pound. 

It  will  be  seen  that  this  method  of  numbering  the  yarn  is 
similar  to  that  explained  in  connection  with  cotton  yarn,  the 
only  difference  being  that  1,600  yards  in  this  case  is  taken  as 
the  standard  in  place  of  840,  which  is  the  standard  for  cotton. 
Consequently,  the  rules  given  for  finding  length,  weight, 
and  counts  of  cotton  yarns  will  be  found  to  apply  equally 
well  in  this  case,  with  the  exception  that  1,600  must  be  used 
in  place  of  840.  However,  in  order  that  the  student  may 


8 


YARN  CALCULATIONS,  GENERAL 


§11 


thoroughly  understand  the  application  of  these  rules,  an 
example  of  each  will  be  given. 


Example  1. — A  woolen  yarn  72,000  yards  in  length  is  found  to 
weigh  10  pounds;  what  run  is  it? 


Solution. — 


72,000 
10  x  1,600 


—  4^-run. 


Ans. 


Example  2. — What  is  the  weight  of  64,000  yards  of  2-run  woolen 
yarn  ? 


Solution.— 


64,000 
2  X  1,600 


20  lb.  Ans. 


Example  3. — What  is  the  length  of  5  pounds  of  a  6-run  yarn? 
Solution. —  5  X  6  X  1,600  =  48,000  yd.  Ans. 


14.  A  very  convenient  method  of  finding  the  weight, 
run,  or  length  of  a  woolen  yarn  when  the  weight  is  expressed 
in  ounces  may  be  explained  as  follows; 

Example  1.— What  is  the  weight,  in  ounces,  of  8,800  yards  of 
a  5i-run  yarn? 

e  8«800 

Solution.—  -  w  ,  ,,nn  =  1  lb. 

5.5  X  1,600 

16  (oz.  in  1  lb.)  X  1  =  16  oz.  Ans. 

It  will  be  noticed  that  as  it  is  desired  to  obtain  the  weight 
in  ounces,  the  result  in  pounds  is  multiplied  by  16,  which 
gives  16  ounces  as  the  answer.  If,  instead  of  multiplying 
the  answer  by  16,  a  new  standard  should  be  obtained 
by  dividing  1,600  by  16,  the  result  would  be  the  same,  but 
the  process  would  be  considerably  shortened.  Therefore, 
whenever  the  weight  is  expressed  in  ounces  the  standard 
employed  should  be  100. 


Example  2. — What  is  the  weight,  in  ounces,  of  5,250  yards  of 
a  3-run  yarn? 

5,250 


Solution. 


3  X  100 


=  17^  oz.  Ans. 


Example  3. —  4,000  yards  of  woolen  yarn  weighs  5  ounces;  what 
run  is  it? 

4,000 


Solution. — 


5  X  100 


=  8-ruu.  Ans. 


Example  4. — How  many  yards  are  there  in  10  ounces  of  a  3-run 
yarn? 

Solution. —  10  X  3  X  100  =  3,000  yd.  Ans. 


§11 


YARN  CALCULATIONS,  GENERAL 


9 


CUT  SYSTEM  OF  NUMBERING 

15.  In  the  vicinity  of  Philadelphia  the  size  of  woolen 
yarns  is  based  on  what  is  termed  the  cut  system,  which 
has  300  yards  for  its  standard  length  of  yarn.  In  all  other 
respects  the  calculations  are  similar;  consequently,  the  rules 
previously  given  will  apply  in  this  case,  with  the  exception 

r 

that  300  is  used  as  the  standard. 

Example  1. — What  is  the  size  in  the  cut  system  of  a  woolen  yarn 
36,000  yards  of  which  weighs  10  pounds? 

„  36,000  10  .  . 

Solution. —  =  12-cut.  Ans. 

10  X  300 

Example  2. — What  is  the  weight  of  66,000  yards  of  a  20-cut  yarn? 
Solution. —  20  X  300  ~  ^us' 

Example  3. — What  is  the  length  of  10  pounds  of  a  24-cut  woolen 
yarn  ? 

Solution. —  24  X  10  X  300  =  72,000  yd.  Ans. 


SIZING  WOOLEN  ROVING  AND  YARN 
16.  Both  woolen  roving  (in  some  mills  and  districts  the 
term  roving  is  corrupted  into  roping )  and  yarn  are  sized 
according  to  the  number  of  runs  or  cuts  in  1  pound.  In  sizing 


both  the  yarn  and  roving,  a  scale  known  as  a  run  or  cut 
scale  is  commonly  used.  This  scale,  shown  in  Fig.  3,  con¬ 
sists  of  a  brass  beam  carried  by  an  upright  standard,  and  so 


10 


YARN  CALCULATIONS,  GENERAL 


§11 


arranged  that  when  a  given  number  of  yards  of  roving  or 
yarn  is  placed  on  the  beam  in  such  a  position  as  to  balance 
(the  beam  being  out  of  balance  when  not  in  use),  the  posi¬ 
tion  of  the  yarn  will  indicate  its  size  on  a  graduated  scale 
on  the  beam.  In  order  that  the  reading  may  be  as  accurate 
as  possible,  the  yarn  or  roving  to  be  weighed  is  twisted  into 
a  small  bunch  and  suspended  by  a  single  thread.  Run  or 
cut  scales  are  usually  made  to  indicate  the  correct  size  from 
50  yards  of  yarn  or  roving,  but  they  may  be  made  to  indicate 
the  size  with  any  desired  number  of  yards.  If  other  than 
the  number  of  yards  for  which  the  scale  is  graduated  is 
used  in  weighing,  the  reading  of  the  scale  must  be  corrected. 
In  order  to  do  accurate  work  the  equilibrium  of  a  run  or 
cut  scale  should  be  adjusted  with  a  known  weight  or  a 
known  size  of  yarn.  This  can  be  accomplished  by  turning 
to  the  right  or  the  left  the  setscrew  in  the  heavy  end  of 
the  beam. 

When  a  mill  is  not  equipped  with  a  run  scale,  the  size  of 
the  yarn  must  usually  be  determined  by  means  of  a  grain 
scale.  In  most  mills  20  yards  of  yarn  or  roving  is  measured 
off  for  sizing,  but  in  other  mills  50  yards,  100  yards,  and 
various  other  lengths  are  used. 

17.  To  find  the  size  of  a  woolen  roving  or  yarn  when 
the  weight  of  a  given  number  of  yards  is  known: 

Rule. — Divide  the  number  of  ya?'ds  zveighed,  multiplied  by 
7,000,  by  the  standard  number  ( 1,600  or  300),  multiplied  by 
the  weight ,  in  grains. 

Example. — If  20  yards  of  woolen  yarn  weighs  25  grains,  what  is 
the  run  of  the  yarn? 

c  20  X  7,000  o1 

Solution.  1|600  x  g,  =  Si-run.  Ans. 

The  following  table  gives  the  size,  in  runs,  and  the  weight, 
in  grains,  of  20  yards  of  woolen  roving  or  yarn.  If  any 
number  of  yards  other  than  20  is  weighed,  the  reading  of  the 
table  will  have  to  be  corrected.  For  instance,  if  100  yards 
of  yarn  is  found  to  weigh  175  grains,  the  size  of  the  yarn  is 
not  2-run,  but  5  times  (100  -f-  20  =  5)  i-run,  or  22-run. 


11 


YARN  CALCULATIONS,  GENERAL 


11 


TABLE  IV 


Runs 

Grains 

Runs 

Grains 

Runs 

Grains 

1 

2 

175.00 

5 

17.50 

10 

8-75 

5 

8 

140.00 

5  4 

16.66 

ioi 

8-53 

3 

4 

I  l6.66 

5* 

15.90 

ioi 

8-33 

7 

8 

100.00 

5f 

15.21 

1  of 

8.13 

i 

87.50 

6 

14.58 

1 1 

7- 95 

I  4 

70.00 

67 

14.00 

1  if 

7-77 

i! 

58-33 

6i 

13.46 

1  is 

7.60 

if 

50.00 

6f 

12.96 

1  if 

7-44 

2 

43-75 

7 

12.50 

12 

7.29 

21 

38.88 

74 

12.06 

I2f 

7.14 

2| 

35-00 

7s 

1 1 .66 

12! 

7.00 

2T 

31.81 

7f 

.  1 1 .29 

I2f 

6.86 

3 

29.16 

8 

10.93 

13 

6-73 

37 

26.92 

r-t|^ 

OO 

10.60 

137 

6.60 

32 

25.00 

8i 

10.29 

132 

6.48 

34 

23-33 

8f 

10.00 

1 3f 

6.36 

4 

21.87 

9 

9.72 

14 

6.25 

44 

20.58 

94 

9-45 

142^ 

6.03 

42 

19.44 

92 

9.2  1 

15 

5-83 

44 

18.42 

97 

8.97 

WORSTED  YARNS 

18.  The  basis  on  which  worsted  yarns  are  numbered  is 
that  of  having  560  yards  constitute  1  hank,  and  numbering 
the  yarns  according  to  the  number  of  hanks  that  weigh 
1  pound;  consequently,  the  student  will  understand  that  the 
rules  given  for  numbering  cotton  yarns  apply  equally  well 
when  numbering  worsted  yarns,  with  the  difference,  how¬ 
ever,  that  560  yards  is  taken  as  the  standard  instead  of  840. 
Examples  for  finding  counts,  weight,  and  length  of  worsted 
yarns  will  be  given  in  order  further  to  enable  the  student  to 
understand  the  system  employed. 

Example  1. —  268,800  yards  of  worsted  yarn  weighs  10  pounds. 
What  are  the  counts? 


12 


YARN  CALCULATIONS,  GENERAL 


11 


0  268,800 

Solution. —  ..  w  =  48s.  Ans. 

10  X  5b0 

Example  2. — What  is  the  weight  of  161,280  yards  of  a  36s  worsted 
yarn? 

161,280  0  . 

Solution.—  =  8  lb.  Ans. 

36  X  560 

Example  3. — What  is  the  length  of  20  pounds  of  a  28s  worsted 
yarn  ? 

Solution. —  20  X  28  X  560  =  313,600  yd.  Ans. 

Besides  the  table  of  weights  given  in  connection  with 
cotton  yarns,  the  following  table  of  lengths  is  used  for 
worsted  yarns: 

TABLE  V 

1  yard  =  1  thread  (circumference  of  reel). 

80  yards  =  80  threads  =  1  lea. 

560  yards  =  560  threads  =  7  leas  =  1  hank. 


SIZING  WORSTED  YARNS 

19.  When  finding  the  counts  of  worsted  yarns  it  is  the 
usual  custom  to  reel  off  80  yards  and  divide  the  weight,  in 
grains,  into  1,000.  It  will  be  seen  that  this  method  is  similar 
to  that  of  reeling  120  yards  of  cotton  and  dividing  its  weight 
into  1,000,  as  previously  explained,  since  one-seventh  of  560 
(standard  length  for  worsted)  is  80.  Other  lengths  and 
dividends  for  worsted  yarns  are  as  follows: 


TABLE  VI 


If  320  yards  is  weighed,  divide 
If  160  yards  is  weighed,  divide 
If  80  yards  is  weighed,  divide 
If  60  yards  is  weighed,  divide 
If  40  yards  is  weighed,  divide 
If  30  yards  is  weighed,  divide 
If  20  yards  is  weighed,  divide 
If  15  yards  is  weighed,  divide 
If  10  yards  is  weighed,  divide 
If  8  yards  is  weighed,  divide 
If  6  yards  is  weighed,  divide 
If  4  yards  is  weighed,  divide 
If  3  yards  is  weighed,  divide 
If  2  yards  is  weighed,  divide 
If  1  yard  is  weighed,  divide 


weight, 

in 

grains, 

into 

4,000. 

weight, 

in 

grains, 

into 

2,000. 

weight, 

in 

grains, 

into 

1,000. 

weight, 

in 

grains, 

into 

750. 

weight, 

in 

grains, 

into 

500. 

weight, 

in 

grains, 

into 

375. 

weight, 

in 

grains, 

into 

250. 

weight, 

in 

grains, 

into 

187|. 

weight, 

in 

grains, 

into 

125. 

weight, 

in 

grains, 

into 

100. 

weight, 

in 

grains, 

into 

75. 

weight, 

in 

grains, 

into 

50. 

weight, 

in 

grains, 

into 

371. 

weight, 

in 

grains, 

into 

25. 

weight, 

in 

grains, 

into 

12*. 

§11 


YARN  CALCULATIONS,  GENERAL 


13 


SIZING  WORSTED  ROVING 

20.  It  is  usually  customary  to  designate  the  size  of 
worsted  roving  by  the  number  of  drams  that  40  yards  weighs; 
thus,  if  40  yards  of  a  certain  roving  weighs  3.2  drams  it  is 
known  as  a  3.2-dram  roving.  When  very  fine  yarns  are 
being  made,  however,  the  roving  is  generally  designated  by 
the  weight,  in  drams,  of  80  yards,  this  length  being  taken 
because  of  the  greater  accuracy  obtained  by  weighing  a 
greater  length.  In  sizing  the  roving,  a  scale  capable  of 
weighing  drams  and  tenths  of  drams  is  commonly  used,  but 
should  it  be  necessary  to  use  a  grain  scale  the  measure¬ 
ments  given  in  Table  II  should  be  employed. 

Since  worsted  yarn  is  numbered  on  the  basis  of  the 
number  of  hanks  of  560  yards  in  1  pound  and  roving  by  the 
number  of  drams  that  40  or  80  yards  weighs,  it  is  sometimes 
desirable  to  find  the  counts  of  a  roving. 

21.  To  find  the  counts  of  a  roving  when  its  weight  in 
drams  is  known: 

Rule. — If  the  weight ,  in  drams ,  of  40  yards  is  known ,  divide 
this  zv eight  fn to  18.3.  If  the  zv eight,  i?i  drams ,  of  80  yards  is 
known ,  divide  this  weight  into  36.6. 

Example. — What  are  the  counts  of  worsted  roving  40  yards  of  which 
weighs  6.1  drams? 

Solution. —  18.3  -f-  6.1  =  3s.  Ans. 

Note. —  18.3  and  36.6  are  known  as  ftauge  points  and  are  obtained  by  multi¬ 
plying  the  drams  in  1  pound  by  the  number  of  yards  weighed  (40  or  80)  and  dividing 
by  the  number  of  yards  in  1  hank;  thus, 

266  V  40 

— — j —  =  18.285,  practically  18.3 
256|360"  =  36‘57,  practically  36-6 

From  this  it  will  be  seen  that  40  yards  o£  number  Is  weighs  18.3  drams,  and 
80  yards  of  number  Is  weighs  36.6  drams,  which  is  the  true  significance  of  the 
gauge  points. 


14 


YARN  CALCULATIONS,  GENERAL 


811 


SILK  YARNS 


22.  Two  classes  of  silk  are  used  for  manufacturing 
purposes — spun  and  raw  silk.  As  these  are  numbered  in  an 
entirely  different  manner,  they  will  be  considered  separately. 

23.  Single  spun  silk  is  numbered  in  a  manner  exactly 
similar  to  cotton  yarns,  containing  840  yards  to  the  hank  and 
basing  the  counts  on  the  number  of  hanks  that  weigh  1  pound. 


Example  1.— What  are  the  counts  of  a  spun-silk  thread,  if  756,000 
yards  weighs  15  pounds? 


Solution.— 


756,000 
15  X  840 


=  60s. 


Ans. 


Example  2. — What  is  the  weight  of  403,200  yards  of  a  40s  spun- 
silk  thread? 


Solution. — 


403,200 
40  X  840 


=  12  lb. 


Ans. 


Example  3. — What  is  the  length  of  4  pounds  of  a  48s  spun-silk 
yarn? 

Solution. —  4  X  48  X  840  =  161,280  yd.  Ans. 


24.  There  are  two  distinct  systems  employed  for  num¬ 
bering  raw  silk,  one  of  which  is  used  for  designating  the 
size  of  raw  silk  as  it  comes  from  the  cocoon,  while  the  other 
is  used  for  numbering  thrown  silk ,  i.  e.,  raw  silk  that  has  been 
doubled  and  reeled. 

Raw  silk  as  it  comes  from  the  cocoon  is  numbered  accord¬ 
ing  to  the  weight  of  one  skein  476  meters  in  length.  This 
weight  is  expressed  in  units  equal  to  -gVth  part  of  a  denier ,  or 
.0531  gram.  Although  .0531  gram  is  really  equal  to  only  a 
fractional  part  of  a  denier,  it  is  commonly  spoken  of  as 
a  denier.  The  so-called  denier  is  actually  equal  to 
.053115  gram,  but  in  practice  .0531  gram  is  used,  while 
476  meters  is  equal  to  520.57  yards,  although  in  practice 
520  yards  is  used  as  the  standard  length  of  a  skein.  If 
therefore  520  yards  of  raw  silk  weighs  .7965  gram  (.0531 
X  15  =  .7965),  it  is  spoken  of  as  a  15-denier  silk.  From 
this  explanation  it  will  be  noted  that  the  system  of  numbering 
raw  silk  differs  materially  from  that  employed  for  numbering 
cotton,  woolen,  worsted,  spun  silk,  etc.,  since  the  higher  the 


11 


YARN  CALCULATIONS,  GENERAL 


15 


number,  the  coarser  is  the  silk;  whereas,  in  the  systems 
previously  explained,  higher  counts  indicate  finer  yarns. 

Example  1. — If  520  yards  of  raw  silk  weighs  .6372  gram,  what  is 
the  size  of  the  silk? 

Solution. —  .6372  -=-  .0531  (grams  in  1  denier)  =  12-denier  silk. 

Ans. 

Example  2. — What  is  the  weight,  in  grams,  of  15,600  yards  of 
20-denier  raw  silk? 

Solution. —  520  yards  weighs  20  deniers,  or  1.062  grams  (.0531 
X  20  =  1.062);  15,600  yards,  therefore,  since  520  is  contained  in  15,600 
exactly  30  times,  will  weigh  30  X  1.062  grams,  or  31.860  grams.  Ans. 

Example  3. — How  many  yards  are  there  in  4.248  grams  of  26-denier 
raw  silk? 

Solution. — Since  520  yards  weighs  26  deniers,  20  yards  (520  -=-  26 
=  20)  weighs  1  denier,  or  .0531  gram.  If  20  yards  weighs  .0531  gram, 
there  will  be  as  many  times  20  yards  in  4.248  grams  as  .0531  is  contained 
in  4.248,  or  80  times  (4.248  -f-  .0531  =  80) .  Therefore,  80  times  20  yards 
=  1,600  yards.  Ans. 

Note.— There  are  practically  28j  grams,  or  533i  deniers,  in  1  ounce  (avoirdupois). 

25.  Although  in  Europe  thrown  silk  is  numbered  the 
same  as  raw  silk,  in  America  a  different  system  is  employed; 
namely,  the  dram  system,  in  which  the  method  of  speci¬ 
fying  the  size  of  such  yarns  is  that  of  giving  the  weight  of 
1,000  yards  in  drams  (avoirdupois).  For  example,  if 
1,000  yards  of  a  certain  thrown-silk  thread  weighs  8  drams, 
it  is  known  as  an  8-dram  silk.  It  will  be  seen  from  this  that 
the  system  employed  in  thrown-silk  numbering  also  differs 
materially  from  that  employed  with  cotton,  woolen,  etc.,  since 
in  this  case,  as  well  as  in  the  case  of  raw  silk  as  it  comes 
from  the  cocoon,  the  higher  the  count  of  the  thread,  the 
coarser  it  will  be. 

When  figuring  the  counts  of  thrown  silk,  Table  II,  given 

in  connection  with  cotton  yarns,  must  be  employed. 

<* 

Example  1.— How  many  yards  are  there  in  1  pound  of  a  5-dram 
silk? 

Solution. —  1,000  -5-  5  =  200  yd.  in  1  dram 

200  X  256  (drams  to  the  pound)  =  51,200  yd.  in  1  lb. 

Ans. 


16  YARN  CALCULATIONS,  GENERAL  §11 

Example  2. — How  many  yards  are  therein  10  pounds  of  a  2^-dram 
silk  thread? 

Solution. —  1,000  -f-  2\  =  400  yd.  in  1  dram 
400  X  256  -  102,400  yd.  in  1  lb. 

102,400  X  10  =  1,024,000  yd.  in  10  lb.  Ans. 

Example  3. — What  are  the  counts  of  a  thrown-silk  thread  if  1  pound 
contains  25,600  yards? 

Solution. —  25,600  -f-  256  =  100  yd.  in  1  dram 

1,000  4-  100  =  10  drams  —  weight  of  1,000  yd. 

Therefore,  the  thread  is  a  10-dram  silk.  Ans. 

Example  4. — What  are  the  counts  of  a  thrown-silk  thread  that  con¬ 
tains  800  yards  in  1  ounce? 

Solution.—  800  4-  16  =  50  yd.  in  1  dram 

1,000  4-  50  =  20  drams  =  weight  of  1,000  yd. 

Therefore,  the  thread  is  a  20-dram  silk.  Ans. 

Example  5. — What  is  the  weight  of  1,000  yards  of  a  10-dram  silk 
thread? 

Solution. — The  weight  is  10  drams,  since  thrown  silk  is  numbered 
by  the  dram  weight  of  1,000  yd.  Ans. 

Example  6. — What  is  the  weight  of  48,000  yards  of  an  8-dram  silk? 

Solution. —  1,000  yd.  weighs  8  drams.  Therefore,  48,000  yd.  will 
weigh  48  X  8  drams  =  384  drams.  Ans. 


JUTE,  LINEN,  ANTI  RAMIE 

26.  Fine  jute  yarns,  fine  linen,  and  ramie,  or  China  grass, 
are  numbered  in  a  manner  exactly  like  the  woolen  cut  sys¬ 
tem;  that  is,  300  yards  to  the  hank  is  taken  as  the  standard 
length,  and  the  number  of  hanks  that  weigh  1  pound  deter¬ 
mines  the  counts  of  the  yarn.  Consequently,  all  calculations 
given  in  connection  with  the  cut  system  of  numbering 
woolen  yarns  will  apply  equally  well  when  dealing  with 
these  materials. 

The  counts  of  coarse  jute  and  coarse  linen  yarns  are  deter¬ 
mined  by  the  weight,  in  pounds,  of  a  spyndle  (14,400  yards). 
Thus,  if  14,400  yards  weighs  4  pounds,  it  is  known  as 
4-pound  yarn.  This  system  will  be  seen  to  compare  with 
the  system  of  numbering  raw  silk,  in  that  the  coarser  the 
yarn,  the  higher  will  be  the  counts;  but,  as  will  be  noticed 


§11 


YARN  CALCULATIONS,  GENERAL 


17 


from  the  preceding-  explanations,  this  method  of  numbering 
yarns  is  the  exception  rather  than  the  rule,  since  generally 
yarns  are  numbered  on  the  basis  of  having  the  higher  counts 
for  the  finer  yarns. 


SUMMARY 

27.  If  the  student  carefully  memorizes  the  following  few 
facts,  he  will  be  saved  a  great  deal  of  difficulty  that  might 
otherwise  occur  when  figuring  any  single  yarns  having  higher 
counts  for  finer  threads.  Such  yarns  are  those  made  from 
cotton,  wool,  worsted,  and  spun  silk,  and  are  the  yarns  that 
will  be  most  frequently  met  with. 

1.  If  the  length  of  yarn  is  given,  always  divide  this  length 
by  the  product  of  the  other  item  given  and  the  standard  to 
find  the  required  item. 

2.  If  it  is  desired  to  find  the  length  of  yarn,  always  multi¬ 
ply  the  given  items  and  the  standard  together. 

3.  Cotton  has  for  its  standard  length  840  yards;  worsted, 
560  yards;  spun  silk,  840  yards;  woolen  (run  system), 
1.600  yards;  woolen  (cut  system),  300  yards. 


EXAMPLES  FOR  PRACTICE 

1.  50,400  yards  of  cotton  yarn  weighs  3  pounds;  what  are  the 

counts?  Ans.  20s 


2.  100,800  yards  of  cotton  yarn  weighs  10  pounds;  what  are  the 

counts?  Ans.  12s 


3.  100  skeins  of  cotton  yarn,  each  containing  4,200  yards,  weighs 

20  pounds;  what  are  the  counts?  Ans.  25s 

4.  What  are  the  cotton  counts  of  yarns  120  yards  of  which  weighs 


respectively  17,  21,  26,  and  32  grains? 


Ans. 


58.82s 

47.61s 

38.46s 

31.25s 


5.  If  a  cop  is  known  to  contain  960  yards  and  35  of  these  cops 
weigh  1  pound,  what  are  the  cotton  counts  of  the  yarn?  Ans.  40s 

6.  Woolen  yarn  weighing  1  pound  contains  18,000  yards;  what  is 

the  number  in  both  run  and  cut  system?  f  11.25-run 

Ans'  160-cut 

7.  What  is  the  weight  of  37,000  yards  of  Is  worsted  yarn? 

Ans.  66.07  lb. 


18 


YARN  CALCULATIONS,  GENERAL 


§11 


8.  What  is  the  weight  of  10,000  yards  of  woolen  yarn,  the  yarn 

being  4-run?  Ans.  25  oz. 

9.  How  many  yards  of  60s  spun-silk  yarn  are  there  in  24  pounds? 

Ans.  1,209,600  yd. 

10.  If  200  yards  of  thrown  silk  weighs  6  drams,  what  number  is  it? 

Ans.  30-dram  silk 

11.  What  is  the  length  of  yarn  in  50  pounds  of  20s  cotton? 

Ans.  840,000  yd. 

12.  A  spool  of  28s  cotton  yarn  weighs  28  ounces.  Find  the  length  of 
yarn  on  the  spool  if  the  spool  itself  weighs  8  ounces.  Ans.  29,400  yd. 

13.  What  is  the  weight  of  336,000  yards  of  40s  spun  silk? 

Ans.  10  lb. 

14.  What  is  the  weight  of  151,200  yards  of  60s  cotton?  Ans.  3  lb. 

15.  How  many  yards  are  there  in  4  ounces  of  5i-run  woolen  yarn? 

Ans.  2,100  yd. 


EQUIVALENT  COUNTS 

28.  Many  calculations  are  met  with  in  which  it  becomes 
necessary  to  place  the  count  of  one  yarn  in  the  system  of 
another.  That  is,  if  a  cotton  thread  is  a  certain  count  in  the 
cotton  system,  it  may  be  necessary  to  learn  what  its  counts 
would  be  if  it  were  numbered  similarly  to  a  worsted  thread. 
When  two,  three,  or  more  threads  made  from  different  raw 
stock  and  numbered  according  to  different  methods  are 
placed  in  the  same  system,  they  are  said  to  be  reduced  to 
equivalent  counts. 

Example. — Suppose  that  it  is  desired  to  learn  what  the  counts  of  a 
15s  cotton  would  be  if  numbered  similarly  to  worsted  yarn. 

Solution. — A  15s  cotton  has  15  X  840  =  12,600  yd.  in  1  lb.  The 
question  then  resolves  itself  into  the  following:  What  are  the  worsted 
counts  of  a  yarn  containing  12,600  yd.  to  the  lb.? 

12,600  -f-  560  =  22.5s,  worsted  counts.  Ans. 

29.  To  find  the  count  of  one  system  that  is  equivalent 
to  that  of  another: 

Rule. — Multiply  the  given  counts  by  the  member  of  yards  hi 
the  standard  length  of  the  specified  system  and  divide  by  the 
number  of  yards  in  the  standard  length  of  the  system  required . 


11 


YARN  CALCULATIONS,  GENERAL 


19 


Example  1. — Find  the  equivalent  of  a  40s  cotton  in  worsted 
counts. 

Solution. —  840  X  40  =  33,600 


33,600  4-  560  =  60s.  Ans. 


Explanation. — Since  there  are  840  yards  of  yarn  in 
1  pound  of  Is  cotton,  there  will  be  40  X  840,  or  33,600,  yards 
in  1  pound  of  40s.  The  question  then  is  to  find  the  worsted 
counts  of  a  yarn  containing  33,600  yards  to  the  pound.  Since 
length  divided  by  (standard  multiplied  by  weight)  equals 
counts,  then  33,600  -4-  (560  X  1)  must  equal  the  counts. 

Example  2. — Find  the  equivalent  of  a  16s  cotton  yarn  in  the  woolen 
run  system. 

Solution. —  840  X  16  =  13,440 

13,440  4-  1,600  =  8.4-run,  woolen.  Ans. 


Example  3. — Place  in  worsted  counts,  a  60s  cotton,  an 
woolen,  and  a  ?0s  worsted. 

Solution. — The  cotton  thread  equals  =  90s  worsted. 

obO 

•  8x1  600 

The  8-run  woolen  thread  equals - =7^ —  =  22.85s  worsted. 

560 

The  20s  worsted  would,  of  course,  be  the  same. 


8-run 

Ans. 

Ans. 


EXAMPLES  FOR  PRACTICE 

1.  Find  the  equivalent  of  a  13s  cotton  in  worsted  counts. 

Ans.  19.5s 

2.  Find  the  worsted  counts  of  a  48s  cotton.  Ans.  72s 

3.  What  are  the  cotton  counts  of  a  40s  worsted?  Ans.  26.66s 

4.  What  counts  of  cotton  are  equivalent  to  90s  worsted?  Ans.  60s 

5.  Convert  7-run  to  cut  system.  Ans.  37.33-cut 

6.  What  are  the  equivalent  counts  of  a  12s  silk  in  woolen  runs? 

Ans.  6.3-run 

7.  Find  the  equivalent  of  a  65s  worsted  in  cotton  counts. 

Ans.  43.33s 


20 


YARN  CALCULATIONS,  GENERAL 


§11 


PLY  YARNS 


METHOD  OF  NUMBERING 

30.  Definition. — Very  frequently  during  the  process 
of  manufacturing  yarns,  two  or  more  threads  are  twisted 
together  to  form  one  coarser  thread.  Such  yarns  are  com¬ 
monly  known  as  ply  yarns,  also  sometimes  called  folded, 
or  twisted,  yarns. 

31.  The  method  of  numbering  cotton  ply  yarns  is  that 
of  giving  the  counts  of  the  single  yarns  that  are  folded  and 
placing  before  these  counts  the  number  that  indicates  the 
number  of  threads  folded;  thus,  2/40s  indicates  that  two 
threads  of  40s  single  yarn  are  folded  together,  the  folded 
yarn  being  equal,  in  weight,  to  a  single  20s  thread.  During 
the  process  of  twisting  a  slight  contraction  takes  place. 
Consequently,  to  make  the  resultant  counts  20s,  the  single 
yarns  that  are  folded  must  necessarily  be  slightly  finer  than 
40s.  However,  this  contraction  will  not  be  considered  in 
the  rules  and  examples  to  be  given,  since  it  is  so  slight  that 
it  is  more  a  matter  of  experience  than  one  of  mathematics. 


CALCULATIONS  OF  COTTON  PLY  YARNS 


FOLDED  YARNS  OF  THE  SAME  COUNTS 

32.  It  is  not  customary  in  mills  to  fold  yarns  of  different 
counts,  since  single  yarns  of  the  same  number  make  the  best 
double,  or  ply,  yarns.  Consequently,  when  yarns  of  the  same 
counts  are  folded,  in  order  to  find  the  counts  of  the  result¬ 
ing  ply  yarn,  it  is  simply  necessary  to  divide  the  counts 
of  the  yarns  folded  by  the  number  of  threads  that  constitute 
the  ply  yarn.  For  example,  if  three  threads  of  90s  cotton 
are  folded  to  form  a  ply  yarn,  the  resultant  yarn  will  be 
equivalent  in  weight  to  a  single  80s  (90  -f-  3  =  30).  The 


YARN  CALCULATIONS,  GENERAL 


21 


§11 


student  should  be  careful  to  distinguish  between  the  counts 
of  the  ply  yarn  and  the  counts  of  the  single  yarn  that  equal 
in  weight  the  ply  yarn.  Thus,  if  three  threads  of  90s  are 
twisted  together,  the  yarn  that  results  from  folding  these 
three  yarns,  or  in  other  words  the  ply  yarn,  is  equal  in 
weight  to  a  single  30s,  although  when  indicating  the  count 
of  the  ply  yarn  it  is  spoken  of  as  a  3-ply  90s. 

The  method  of  finding  the  counts,  weight,  and  length  of 
ply  yarns  is  similar  to  that  explained  in  connection  with 
single  yarns,  with  the  exception  that  the  counts  of  the  ply 
yarn  do  not  indicate  the  actual  counts  of  the  thread  but 
instead  indicate  the  counts  of  the  yarns  folded.  Conse¬ 
quently,  when  figuring  to  find  these  particulars,  the  actual 
weight  of  the  ply  yarn  must  be  taken  into  consideration 
and,  on  this  account,  the  counts  of  the  single  yarn  that 
the  ply  yarn  equals  are  considered  and  not  the  counts  of 
the  single  yarns  that  are  folded. 

Example  1. — What  is  the  weight  of  642,000  yards  of  2-ply  40s  cotton 
yarn  ? 

c  042,000  QO  01  n  . 

Solution. —  ■■■  =  08.21  lb.  Ans. 

LJu  X  o4U 

Explanation. — To  make  a  2-ply  40s,  two  ends  of  40s  are 
twisted  together;  consequently,  a  yard  of  the  ply  yarn  will 
weigh  just  twice  as  much  as  a  yard  of  one  of  the  single 
yarns  folded,  which  will  make  the  ply  yarn  equal  in  weight 
to  a  20s  single  yarn.  Therefore,  20,  which  is  the  actual 
counts  of  the  ply  yarn,  is  used  in  the  calculation.  Since 
length  divided  by  (counts  multiplied  by  standard)  equals 
weight,  then  642,000  -f-  (20  X  840)  must  equal  the  weight  of 
the  yarn. 

Example  2. — What  is  the  length  of  20  pounds  of  2-ply  36s  cotton? 

Solution. —  20  X  18  X  840  =  302,400  yd.  Ans. 

Explanation. —  2-ply  36s  is  composed  of  two  threads 
of  36s  folded  together;  consequently,  the  weight  of  a  yard 
of  the  ply  yarn  must  be  just  twice  that  of  a  yard  of  one  of 
the  ends  folded  to  make  the  ply  yarn.  This  will  make  the 
ply  yarn  equal  in  weight  to  an  18s  single  yarn  and  18s  must 


22 


YARN  CALCULATIONS,  GENERA! 


§11 


be  used  as  the  counts  of  the  ply  yarn  in  the  calculation. 
Since  weight  times  counts  times  standard  equals  length, 
then  20  X  18  X  840  must  equal  the  number  of  yards  in 
20  pounds  of  2-ply  36s. 

Example  3. — What  are  the  counts  of  a  2-ply  cotton  yarn,  352,800 
yards  of  which  weighs  10  pounds? 

0  352,800  ,00,01 

Solution. —  v ,  nAr.  =  42  =  2-ply  84s.  Ans. 

10  X  840  r  7 

Explanation. — Since  length  divided  by  (weight  times 
standard)  equals  counts,  then  352,800  -s  (10  X  840)  must  give 
the  actual  counts  of  the  ply  yarn;  that  is,  this  result  gives  the 
counts  of  the  ply  yarn  considered  as  a  single  yarn,  but  since 
two  single  yarns  are  folded  and  each  of  these  is  just  half  as 
heavy  as  the  folded  yarn,  then  two  ends  of  84s  must  be 
folded  to  make  the  ply  yarn,  which,  consequently,  will  be 
known  as  a  2-ply  84s. 


FOLDED  YARNS  OF  DIFFERENT  COUNTS 

33.  Although  not  a  common  practice,  in  some  cases, 
especially  when  it  is  desired  to  make  a  fancy  yarn,  two 
yarns  of  different  counts  are  folded  and  sometimes  two 
yarns  of  different  materials. 

Suppose,  for  illustration,  that  it  is  desired  to  find  the 
resultant  counts  of  a  40s  cotton  folded  with  a  20s  cotton. 
Take  as  a  basis  840  yards  of  each  yarn;  then  840  yards  of 
the  40s  weighs  pound;  840  yards  of  the  20s  weighs 
To  pound.  Consequently,  after  these  yarns  are  folded,  there 
will  be  840  yards  of  a  ply  yarn  the  weight  of  which  is 
4~o  -f-  2V  =  4at  pound. 

The  example  now  resolves  itself  into  the  following:  What 
are  the  counts  of  a  yarn  840  yards  of  which  weighs  -40  pound? 
Since  length  divided  by  (weight  times  standard)  equals 

counts,  then,  3  ^0  =  13.33s,  counts  of  the  ply  yarn. 

To  X  840 

This  example  has  been  worked  out  to  some  length  in  order 
to  enable  the  student  thoroughly  to  understand  the  method 
of  numbering  ply  yarns.  A  somewhat  shorter  method,  how¬ 
ever,  is  as  follows: 


§11  YARN  CALCULATIONS,  GENERAL  23 

34.  To  find  the  resultant  count  when  two  threads  of 
different  numbers  are  folded: 


Rule. — Multiply  the  two  counts  together  and  divide  the 
result  thus  obtained  by  the  siwi  of  the  counts. 


Example. — Same  as  previous  example. 


Solution. — 


40  X  20 
40  +  20 


13.33s,  counts. 


Ans. 


PLY  YARNS  COMPOSED  OF  MORE  THAN  TWO  THREADS 

35.  In  many  cases  it  will  be  found  necessary  to  find  the 
counts  of  a  ply  yarn  made  from  more  than  two  single  threads. 
Under  such  circumstances  it  will  be  necessary  to  follow  a 
somewhat  different  process.  For  example,  suppose  that 
three  single  threads — 24s,  36s,  and  72s,  respectively — are 
folded  to  form  a  ply  yarn  and  it  is  required  to  ascertain  the 
counts  of  the  resultant  yarn.  This  may  be  done  by  follow¬ 
ing  the  rule  previously  given  and  performing  two  operations 
as  follows: 

First  find  the  counts  of  the  yarn  that  would  result  from 

folding  the  24s  with  the  36s  as  follows:  ^  —  ^  _  14  4S 

24  +  36 

The  example  then  resolves  itself  into  the  following: 
What  are  the  counts  of  a  ply  yarn  made  from  one  thread 

of  14.4s  and  one  of  72s?  14.4  X  72  _  ^s.  Ans. 

14.4  +  72 

A  somewhat  shorter  method  than  this,  however,  may  be 
applied  to  3  or  more  ply  yarns  made  from  different  counts. 

Example. — Same  as  given  above. 

Solution. —  72  -p  72  =  1 

72  -f-  36  =  2 
72  -p  24  =  3 
6 

72  -f-  6  =  12s.  Ans. 

Rule. — Take  the  highest  counts  and  divide  it  by  itself  and  by 
each  of  the  other  counts.  Add  the  results  thus  obtained  and 
divide  this  result  into  the  highest  counts. 


24  YARN  CALCULATIONS,  GENERAL  §11 


36.  To  find  the  resultant  counts  when  more  than  one 
end  of  the  different  counts  are  folded: 

Rule. — Divide  the  highest  counts  by  itself  and  by  each  of  the 
other  counts.  Multiply  the  result  in  each  case  by  the  number  of 
ends  of  that  counts.  Add  the  results  thus  obtained  and  divide 
this  result  into  the  highest  counts. 

Example. —  4  ends  of  80s  and  3  ends  of  60s  are  folded  to  form  a 
ply  yarn;  what  are  the  resultant  counts? 

Solution. —  '  80-5-80  =  1;  1X4  =  4 

80  -5-  60  =  1|;  \\  X  3  =  4 

8 

80  -5-  8  =  10s,  resultant  counts.  Ans. 


37.  When  dealing  with  ply  yarns  it  often  becomes  neces¬ 
sary  to  find  the  counts  of  a  yarn  to  be  folded  with  another 
to  produce  the  given  counts. 


Rule. — Miiltiply  the  two  counts  together  and  divide  by  their 
difference. 


Example. — What  counts  must  be  folded  with  a  50s  to  produce 
a  ply  yarn  equal  in  weight  to  a  30s? 


Solution.— 


50  X  30 
50  -  30 


75s.  Ans. 


Proof. — What  are  the  counts  of  a  ply  yarn  made  by 

twisting  a  50s  with  a  75s?  =  30s.  Ans. 

50  +  75 

38.  Another  calculation  that  must  frequently  be  made 
when  dealing  with  ply  yarns  is  that  of  finding  the  required 
weight  of  each  thread  folded  in  order  to  produce  a  required 
weight  of  the  ply  yarn. 


Rule. — Find  the  counts  resulting  from  folding  the  two  or 
more  threads;  then ,  as  the  counts  of  one  thread  is  to  the  result¬ 
ing  counts,  so  is  the  total  weight  to  the  weight  required  of  that 
thread. 

Example. — It  is  desired  to  produce  100  pounds  of  a  ply  yarn  com¬ 
posed  of  an  80s  and  a  32s  twisted  together;  what  will  be  the  required 
weight  of  the  80s  and  also  of  the  32s? 


11 


YARN  CALCULATIONS,  GENERAL 


25 


80  X  32 

Solution. —  ^  =  22.85s,  counts  of  the  ply  yarn  expressed  in 


single  numbers. 


x  = 


80  +  32 

32  :  22.85  =  100  :  x 
100  X  22.85 


32 


=  71.40  lb.  of  32s.  Ans. 


x  — 


80  :  22.85  =  100  :  x 
100  X  22.85 


80 


=  28.56  lb.  of  80s.  Ans. 


In  a  case  similar  to  the  example  given  above,  after  the 
weight  of  one  thread  has  been  obtained,  it  is  of  course  only 
necessary  to  subtract  that  weight  from  the  total  weight  in 
order  to  obtain  the  weight  of  the  other  thread;  or,  in  case 
more  than  two  threads  are  folded,  then  the  weight  of  one  of 
these  threads  may  always  be  obtained  by  subtracting  the 
combined  weight  of  the  other  threads  from  the  total  weight 
of  the  ply  yarn. 

Note. — In  the  previous  example  the  weight  of  the  80s  yarn  plus  the 
weight  of  the  32s  yarn  should  equal  the  weight  of  the  ply  yarn,  but 
owing  to  the  use  of  decimals,  examples  of  this  kind  seldom  give  exact 
results.  Thus,  71.40  pounds  +  28.56  pounds  =  99.96  pounds;  whereas 
the  total  weight  should  be  100  pounds. 


CALCULATION  OF  COST  OF  PLY  YARNS 

39.  If  the  price  of  each  yarn  is  given  and  it  is  required 
to  find  the  price  per  pound  of  the  resultant  yarn,  it  becomes 
necessary  to  multiply  the  weight  of  each  count  of  yarn  by  its 
price,  add  the  results,  and  divide  by  the  total  weight.  The 
answer  will  be  the  price  per  pound  of  the  ply  yarn. 

Example. — If  in  the  example  given  in  Art.  38,  the  80s  yarn  is 
worth  72  cents  per  pound  and  the  32s  is  worth  48  cents  per  pound,  what 
will  be  the  cost  per  pound  of  the  ply  yarn? 

Solution. — 

71.40  lb.  of  32s  at  48  cents  per  lb.  =  $34.27,  cost  of  the  32s  yarn 

28.56  lb.  of  80s  at  72  cents  per  lb.  =  $20.56,  cost  of  the  80s  yarn 

$34.27  +  $20.56  =  $54.83,  total  cost  of  ply  yarn 

$54.83  -f-  100  =  54.8  cents  per  lb.,  cost  of  the  ply  yarn.  Ans. 

40.  Another  rule  for  finding  the  price  of  2-ply  yarns  when 
the  threads  to  be  twisted  together  are  of  different  values  and 
different  counts  is  as  follows: 


26 


YARN  CALCULATIONS,  GENERAL 


§11 


Rule. — Multiply  the  highest  coiints  by  the  Price  of  the  lowest 
counts  and  the  lowest  counts  by  the  price  of  the  highest.  Add  the 
results  thus  obtained  and  divide  this  result  by  the  sum  of  the 
counts.  The  ansiver  will  be  the  price  of  the  ply  yarn. 

Example. — A  32s  yarn  costs  42  cents  per  pound  and  a  16s  yarn 
costs  18  cents  per  pound;  what  will  be  the  cost  per  pound  of  a  ply 
yarn  resulting  from  twisting  these  two? 

Solution. —  32  X  $.18  =  $5.76;  16  X  $.42  =  $6.72 
$5.76  +  $6.72  =  $12.48;  32  +  16  =  48 
$12.48  4-  48  =  26  cts.  Ans. 


WOOLEN  AND  WORSTED  PLY  YARNS 

41.  Ply  yarns  made  from  worsted  or  woolen  yarns  are 
numbered  exactly  the  same  as  cotton  ply  yarns.  Therefore, 
the  rules  given  will  be  found  to  apply  equally  well  when 
dealing  with  ply  yarns  made  of  worsted  and  woolen,  with 
the  exception  that  in  each  case  the  standard  number  of  yards 
to  the  pound  of  the  system  being  dealt  with  must  be  used. 
Consequently,  ply  yarns  of  these  systems  need  no  further 
explanation  here. 


PLY  YARNS  OF  SPUN  SILK 

42.  The  numbering  of  ply  yarns  made  from  spun  silk 
will  be  found  to  differ  somewhat  from  the  methods  pre¬ 
viously  explained.  Thus,  when  numbering  silk  ply  yarns, 
the  counts  resulting  after  folding  the  yarns  is  given  and  this 
number  is  followed  by  the  number  that  indicates  how  many 
threads  are  folded. 

For  example,  60/2  spun  silk  indicates  that  two  threads 
of  120s  have  been  folded  together.  Thus,  it  will  be  seen 
that  the  actual  counts  of  the  ply  yarn  are  given  instead  of 
the  counts  of  the  single  yarn,  as  is  the  case  in  cotton,  woolen, 
and  worsted  ply  yarns. 

Example  1. — What  is  the  weight  of  642,000  yards  of  a  40s  2-ply 
spun  silk? 

„  642,000  in  in_  ,,  . 

Solution. —  — —  19.10/  lb.  Ans. 

40  X  840 


11 


YARN  CALCULATIONS,  GENERAL 


27 


Explanation.—  40s  2-ply  spun  silk  is  equal  in  weight 
to  a  single  thread  of  40s.  Consequently,  40  should  be  con¬ 
sidered  as  the  counts  of  the  ply  yarn  when  finding  weight  or 
length.  Since  length  divided  by  (counts  times  standard) 
642,000 

equals  weight,  40  x  g40  must  equal  the  weight  of  the  yarn. 

Example  2. — What  is  the  length  of  20  pounds  of  a  30s  2-ply  spun 
silk? 

Solution. —  840  X  30  X  20  =  504,000  yd.  Ans. 

Explanation. — A  30s  2-ply  spun  silk  is  equal  in  weight 
to  a  single  30s;  consequently,  30  should  be  considered  as  the 
counts  of  the  ply  yarn.  Since  standard  times  counts  times 
weight  equals  length,  840  X  30  X  20  must  equal  the  length 
of  the  yarn. 

Example  3. — What  are  the  counts  of  a  2-ply  silk  yarn  if  352,800  yards 
weighs  10  pounds? 

o  352,800  _  , 

Solution. —  1^X840  =  Ans- 

Explanation. — The  counts  of  the  2-ply  yarn  would  be 
indicated  as  follows:  42/2  spun  silk,  which  shows  that 
two  ends  of  84s  have  been  twisted  to  make  the  ply  yarn. 


PLY  YARNS  OF  DIFFERENT  MATERIALS 

43.  In  all  cases  where  threads  of  different  materials  are 
twisted  together,  in  order  to  perform  any  of  the  calculations 
previously  explained,  it  becomes  necessary  first  to  place 
these  counts  in  the  same  system  of  numbering  yarns. 

Example. — A  36s  cotton  and  a  48s  worsted  are  twisted  to  form  a 
ply  yarn;  what  are  the  counts  of  the  resultant  yarn? 

Solution. — It  is  first  necessary  to  ascertain  in  which  system  the 
resultant  yarn  should  be  placed.  In  this  case  the  counts  of  the  ply 
yarn  will  be  found  in  both  the  worsted  and  cotton  systems.  In  the 
first  case  then,  to  find  the  worsted  counts  of  the  ply  yarn  resulting 
from  twisting  these  two  yarns  it  is  necessary  to  find  the  equivalent 

36  X  840 


counts  of  the  36s  cotton  in  the  worsted  system. 


560 


2.S 


YARN  CALCULATIONS,  GENERA! 


11 


The  36s  cotton  is  found  to  equal  a  54s  worsted  so  that  the  question 
resolves  itself  into  the  following:  What  are  the  counts  of  a  ply  yarn 
resulting  from  twisting  a  54s  worsted  and  a  48s  worsted? 

54  X  48 

1-- - =  25.41,  worsted  counts  of  the  ply  yarn.  Ans. 

54  +  48 

Since  in  this  example  it  is  also  required  to  find  the  counts  of  the 
ply  yarn  in  the  cotton  system,  it  is  therefore  necessary  first  to  find 

48  X  560 


the  equivalent  counts  of  the  48s  worsted  in  cotton. 


840 


=  32s. 


Having  placed  the  48s  worsted  in  the  cotton  system,  treat  the 
worsted  as  though  it  were  cotton  and  find  the  counts  of  a  ply  yarn 
that  will  result  from  folding  a  32s  and  a  36s  cotton. 

32  X  36 

.  rr  =  16.94,  cotton  counts  of  the  ply  yarn.  Ans. 

oZ  «j o 

From  this  it  is  seen  that  if  a  36s  cotton  and  a  48s  worsted  are 
twisted  together,  the  counts  of  the  resultant  ply  yarn  will  be  either 
25.41s  worsted  or  16.94s  cotton. 


EXAMPLES  FOR  PRACTICE 

1 .  If  a  thread  of  40s  cotton  and  one  of  60s  cotton  are  twisted 
together,  what  are  the  counts  of  the  resultant  ply  yarn?  Ans.  24s 

2.  A  20s,  30s,  and  60s  cotton  are  twisted  together;  what  are  the 

counts  of  the  ply  yarn?  Ans.  10s 

3.  What  are  the  counts  of  a  yarn  that  must  be  twisted  with  a 

44s  cotton  to  produce  a  ply  yarn  equal  to  a  20s?  Ans.  36.66s 

4.  What  weight  of  40s  cotton  yarn  must  be  twisted  with  a  20s  to 

produce  200  pounds  of  ply  yarn?  Ans.  66.65  lb.  of  40s 

5.  What  weight  of  100s,  80s,  and  60s  cotton  would  be  used  in  pro¬ 
ducing  500  pounds  of  ply  yarn?  f  127.65  lb.  of  100s 

Ans.<  159.56  lb.  of  80s 
[212.75  lb.  of  60s 

6.  What  counts  of  ply  yarn  would  result  if  18s,  24s,  and  30s  cotton 

are  twisted  together,  and  what  weight  of  each  would  there  be  in 
240  pounds  of  the  folded  yarn?  f  106. 66f  lb.  of  18s 

Ans.  8s I  80.00  lb.  of  24s 
[53.33ilb.  of  36s 

7.  A  ply  yarn  is  made  from  one  end  each  of  10s,  12s,  and  15s;  what 

are  the  counts  of  the  ply  yarn?  Ans.  4s 

8.  Find  the  price  per  pound  and  the  weight  of  each  yarn  in 
100  pounds  of  a  ply  yarn  made  from  one  thread  of  40s  2-fold  silk  at 
$2.52  per  pound  and  one  thread  of  4-run  woolen  at  40  cents  per  pound. 

173.92  ct.  per  pound 
Ans.  <  16  lb.  of  40s-2  silk 

[84  lb.  of  4-run  woolen 


§11 


YARN  CALCULATIONS,  GENERAL 


29 


9.  A  3-ply  yarn  is  made  from  80s,  40s,  and  30s  worsted  and  weighs 

100  pounds;  what  weight  does  it  contain  of  each  count  of  yarn  and 
what  are  the  counts  of  the  ply  yarn?  ( 17.64  lb.  of  80s 

Ans.  14. 117s<  35.29  lb.  of  40s 
[47.05  lb.  of  30s 

10.  What  is  the  cost  per  pound  of  a  ply  yarn  composed  of  one 

thread  of  22s  yarn  at  40  cents  per  pound  and  one  thread  of  40s  at 
84  cents  per  pound?  Ans.  55.6  ct. 

11.  What  is  the  price  per  pound  of  a  ply  yarn  composed  of  one 

thread  of  40s  worsted  at  96  cents  per  pound  and  one  thread  of  80s  2-fold 
silk  at  $5.28  per  pound?  Ans.  $2.04 

12.  A  ply  yarn  is  composed  of  one  thread  of  10s  cotton  and  one 
thread  of  26-cut  woolen;  what  are  the  counts  in  the  cut  system? 

Ans.  13.481-cut 

13.  What  would  be  the  resultant  counts  in  the  cotton  system  of  one 

end  of  40s  cotton,  one  of  40s  worsted,  and  one  of  60s  silk  twisted 
together?  Ans.  12.631s  cotton 

14.  What  counts  result  from  twisting  a  60s  cotton  with  a  36s 
worsted?  Give  answers  in  both  cotton  and  worsted  systems. 

,  f  17.14s  in  cotton  system 
s‘  125.71s  in  worsted  system 


DIAMETER  OF  YARNS 


METHODS  OF  DETERMINING  THE  DIAMETER 

OF  YARNS 

44.  Definition.— By  the  term  diameter  of  yarns  is 
meant  their  thickness.  Thus*  if  a  thread  is  said  to  be 
gV  inch  in  diameter,  it  means  that  it  is  -gV  inch  in  thickness, 
or  in  other  words,  that  sixty  of  these  threads  can  be  placed 
side  by  side  in  1  inch.  The  diameters  of  threads  become  an 
important  matter  when  considering  the  number  of  threads 
that  should  be  placed  in  a  given  space  of  cloth,  since  if  too 
small  a  number  of  ends  are  put  in  a  cloth,  it  will  not  in  most 
cases  give  good  results,  while,  on  the  other  hand,  if  more 
ends  are  crowded  into  a  given  space  than  can  lie  side  by 
side  in  that  space,  it  will  give  the  cloth  a  cockled  appear¬ 
ance.  There  are  two  universal  methods  of  finding  the 
diameters  of  yarns — by  means  of  the  microscope  and  a 


30 


YARN  CALCULATIONS,  GENERAL 


§11 


micrometer,  and  by  means  of  the  following  rules;  but  it 
should  be  understood  that  these  rules  give  only  approximate 
results,  since  the  diameter  of  a  thread  is  influenced  very 
largely  by  the  class  of  material  used  and  by  the  twist,  per 
inch  placed  in  the  thread. 

45.  A  rule  for  finding  the  diameter  of  yarn  that  is  based 
very  largely  on  experience  is  as  follows: 

Rule. — Find  the  number  of  yards  per  pound  in  the  counts 
under  consideration  a?id  extract  the  square  root  of  this  number. 
This  result  less  a  deductio)i  gives  the  denominator  of  a  fraction 
having  1  as  its  numerator  that  expresses  the  diameter  of  the  yarn 
in  inches. 

The  deductions  generally  made  are  16  per  cent,  for 
woolen,  10  per  cent,  for  worsted,  and  8  per  cent,  for  cotton. 

Example. — Find  the  diameter  of  a  40s  worsted  thread. 

Solution. —  40  X  560  =  22,400  yd.  per  lb. 

>122,400  =  150,  nearly 
150  less  10  per  cent.  =  135 

Therefore,  the  diameter  of  a  40s  worsted  is  approximately  rss  in-, 
or  in  other  words,  135  ends  of  a  40s  worsted  can  be  placed  side  by 
side  in  1  in.  Ans. 


BEAMED  YARNS 

46.  A  part  of  the  yarn  before  being  woven  into  cloth  is 
placed  on  what  are  known  as  loom  beams,  a  large  number 
of  ends  of  the  same  length  being  placed  on  one  beam.  The 
calculations  necessary  in  connection  with  the  yarn  on  a  beam 
will  be  found  to  be  very  similar  to  those  used  in  connection  with 
the  length,  weight,  and  counts  of  single  ends  with  this  differ¬ 
ence;  viz.,  that  whereas  formerly  only  a  single  end  was  being 
dealt  with,  in  the  present  case  a  large  number  of  ends  must 
be  taken  into  consideration.  Thus,  for  example,  if  each  end 
on  a  beam  is  1,000  yards  long  and  there  are  2,000  ends,  then 
there  must  be  2,000  X  1,000  =  2,000,000  yards  of  yarn. 
This  point  should  always  be  taken  into  consideration  when 
dealing  with  yarn  placed  on  a  beam.  The  yarn  on  a  beam 
is  generally  spoken  of  as  the  warp  yarn,  or  the  warp. 


11 


YARN  CALCULATIONS,  GENERAL 


31 


Note. — Since  the  examples  and  rules  for  finding  different  data  in 
regard  to  the  yarn  on  a  beam  apply  equally  as  well  to  one  system  of 
numbering  yarns  as  to  another,  no  particular  system  will  be  dealt  with, 
the  examples  being  taken  from  different  classes  of  yarn. 

47.  To  find  the  counts  of  the  yarn  on  a  beam  containing 
only  one  size  of  yarn,  the  weight,  length,  and  number  of 
ends  being  given: 

Rule. — Multiply  the  length ,  expressed  in  yards ,  by  the  num¬ 
ber  oh  ends  on  the  beam  and  divide  the  result  thus  obtained  by  the 
weight ,  expressed  in  pounds ,  times  the  sta?idard  number  oh  yards 
to  the  pound. 

Example. — A  warp  beam  contains  2,400  ends  of  cotton  each  200 
yards  long.  The  weight  of  this  yarn  is  15  pounds;  what  are  the  counts? 

o  200  X  2,400  oc  . 

Solution.—  -  =  38.095s.  Ans. 

lo  X  o4u 

Explanation. — Since  there  are  2,400  ends  and  each  end 
is  200  yards  long,  there  must  be  2,400  X  200  =  480,000  yards 
in  all.  The  question  then  resolves  itself  into  finding  the 
counts  of  a  yarn  480,000  yards  of  which  weighs  15  pounds. 
Since  length,  in  yards,  divided  by  (weight,  in  pounds,  times 
standard)  always  equals  counts,  480,000  divided  by  (15  X  840) 
must  give  the  counts. 

In  some  cases  the  weight  given  will  be  found  to  include 
not  only  the  weight  of  the  yarn,  but  also  that  of  the  beam  on 
which  the  yarn  is  placed.  When  this  occurs,  it  is  necessary 
first  to  deduct  the  weight  of  the  beam  from  the  weight  given, 
in  order  to  obtain  the  true  weight  of  the  yarn. 


48.  To  find  the  number  of  ends  on  a  beam  when  weight, 
length  of  the  warp,  and  size  of  the  yarn  are  known: 


Rule. — Multiply  the  weight ,  in  pounds ,  by  the  standard 
number  and  by  the  size  oh  the  yarn.  Divide  the  result  thus 
obtained  by  the  length  oh  the  warp ,  in  yards. 


Example  1. — A  worsted  warp  is  800  yards  long  and  weighs 
200  pounds  exclusive  of  the  beam.  If  the  warp  is  composed  of 
20s  yarn,  how  many  ends  does  it  contain? 


200  X  560  X  20 


2,800  ends.  Ans. 


Solution. 


800 


32 


YARN  CALCULATIONS,  GENERAL 


§11 


Example  2. — A  woolen  warp  is  600  yards  long  and  weighs 
120  pounds  exclusive  of  the  beam.  If  the  warp  is  composed  of 
4|-run  yarn,  how  many  ends  does  it  contain? 

120  X  1,600  X  41 


Solution. — 


600 


- =  1,440  ends.  Ans. 


49.  To  find  the  weight  of  yarn  on  a  beam  when  length, 
number  of  ends,  and  counts  are  given: 


Rule. — Multiply  the  length ,  expressed  in  yards ,  by  the  num¬ 
ber  oh  ends  on  the  beam  and  divide  the  result  thus  obtained  by 
the  standard  number  of  yards  times  the  counts  of  the  yarn. 


Example. — A  beam  contains  2,400  ends  of  20s  cotton,  each  end 
being  500  yards  long;  find  the  weight  of  the  yarn. 


Solution. — 


500  X  2,400 
840  X  20 


71.428  lb.  Ans. 


Explanation. — By  multiplying  the  length  of  one  end  by 
the  total  number  of  ends  on  the  beam  the  total  length  of 
yarn  on  the  beam  is  obtained;  and  since  the  length,  expressed 
in  yards,  divided  by  the  standard  times  the  counts  equals 
the  weight,  in  pounds  (2,400  X  500)  -4-  (840  X  20),  will  give 
the  weight  of  the  yarn  on  the  beam. 

50.  To  find  the  length  of  a  warp  when  weight,  number 
of  ends,  and  size  of  the  yarn  are  known: 

Rule. — Multiply  the  weight  of  the  warp ,  in  pounds ,  by  the 
standard  number  and  by  the  size  of  the  yarn ,  and  divide  the 
result  thus  obtained  by  the  number  of  ends  in  the  warp. 

Example  1. — A  worsted  warp  contains  2,400  ends  of  12s  yarn  and 
weighs  200  pounds;  how  long  is  it? 

„  200  X  560  X  12  C/VA  J  A 

Solution.—  - 2~40b - =  •^ns> 

Example  2. — A  woolen  warp  contains  1,200  ends  of  2-run  yarn  and 
weighs  231  pounds.  How  long  is  it? 

„  231  X  1,600  X  2  J  A 

Solution. — - —*>00 -  =  616  yd.  Ans. 

51.  To  find  the  length  of  warp  that  can  be  placed  on 
a  beam: 


Rule. — Find  the  weight  of  yarn  that  the  beam  will  contain , 
by  weighing  a  beam  of  the  same  size  when  filled  with  yarn  and 


§11 


YARN  CALCULATIONS,  GENERAL 


33 


deducting  the  weight  of  the  bea?n  itself.  Then  apply  the  rule  in 

Art.  50. 


Example  1. — A  certain  size  beam  when  filled  with  yarn  weighs 
140  pounds,  the  beam  itself  weighing  50  pounds.  What  length  of  a 
warp  composed  of  1,800  ends  of  20s  cotton  can  be  placed  on  it? 


Solution. — 


140  —  50  =  90  lb.  of  yarn 


90  X  840  X  20 
1,800 


840  yd.  Ans. 


Example  2. — Same  beam  as  in  example  1 .  How  many  yards  of 
a  warp  containing  3,000  ends  of  40s  cotton  can  be  placed  on  it? 


Solution.— 


140  —  50  =  90  lb.  of  yarn 


90  X  840  X  40 
3,000 


1,008  yd.  Ans. 


AVERAGE  NUMBERS 

52.  In  case  different  counts  of  yarn  are  placed  on  the 
same  beam,  as  very  frequently  occurs,  it  will  be  found  neces¬ 
sary  to  first  find  the  average  number ,  or  average  counts,  of  the 
different  yarns  before  making  other  calculations.  By  the 
term  average  number,  or  average  counts,  is  meant  a 
count  of  yarn  that  will  give  the  same  weight,  provided  that 
the  same  number  of  ends  and  the  same  length  occur  in  both 
cases.  Thus,  if  400  ends  of  10s  and  800  ends  of  20s  weigh  a 
certain  number  of  pounds,  then  1,200  (400  +  800)  ends  of  the 
average  counts  will  weigh  the  same,  provided  that  the  ends 
are  the  same  length  in  both  cases. 

53.  To  find  the  average  counts  of  the  ends  on  a  beam 
when  the  ends  are  of  different  counts: 

Rule. — Divide  the  total  number  of  ends  of  each  count  by  its 
own  count.  Add  these  results  together  and  divide  the  result  thus 
obtained  into  the  total  number  of  ends  in  the  warp. 

Example. — There  are  placed  on  the  same  beam  1,800  ends  of  60s 
cotton  and  800  ends  of  40s  cotton;  what  are  the  average  counts? 

Solution. —  1  8  0  0  =  60  =  30 

800  =  40  =  20 
2600  50 

2,600  -r-  50  =  52s,  average  counts.  Ans. 


34 


YARN  CALCULATIONS,  GENERAL 


§11 


Explanation. — Suppose  that  each  end  on  the  beam  is 
,  840  yards  long,  then  (1,800  X  840)  4-  (60  X  840)  =  30  pounds, 

.  weight  of  the  60s  yarn.  Also  (800  X  840)  4-  (40  X  840) 
=  20  pounds,  weight  of  the  40s.  The  combined  weight  of 
the  60s  and  40s  will,  of  course,  be  30  +  20  =  50  pounds. 
There  are  2,600  ends  each  840  yards  long  and  this  length  of 
yarn  weighs  50  pounds.  The  question,  therefore,  is  if  all  the 
yarn  is  the  same  counts,  what  must  be  the  counts  when 
2,600  X  840  yards  weighs  50  pounds?  Since  length,  expressed 
in  yards,  divided  by  standard  times  weight,  expressed  in 
pounds,  equals  counts,  then  (2,600  X  840)  4-  (840  X  50)  must 
equal  the  average  counts.  Since  the  length  in  this  case,  or 
840  yards,  cancels  in  every  calculation  with  the  standard 
number  of  yards,  which  is  also  840,  these  two  terms  may  be 
omitted  and  the  example  stated  as  in  the  solution.  This 
holds  good  with  any  length. 

54.  In  case  more  than  two  different  counts  are  placed 
on  the  same  beam,  the  same  rule  will  be  found  to  apply. 

Example. — What  are  the  average  counts  in  case  200  ends  of  20s, 
1,000  ends  of  40s,  and  900  ends  of  45s  are  placed  on  the  same  beam? 

Solution. —  2  0  0  4-  20  =  10 

10004- 40  =  25 
9004- 45  -  20 
2T(To  5~5 

2,100  4-55  =  38.18s,  average  counts.  Ans. 

55.  In  cases  where  the  order  of  arranging  the  different 
counts  of  yarn  in  the  warp  is  given,  the  total  number  of 
ends  in  the  warp  not  being  known,  the  same  rule  will  be 
found  to  apply  by  considering  the  number  of  ends  in  the 
arrangement  as  the  total  number  of  ends. 

Example. — A  warp  is  arranged  48  ends  of  36s  and  2  ends  of  10s; 
find  the  average  number. 

Solution. —  4  8  4-  36  =  1.3  3  3 

2  4- 10  =  .2 

5  0  1.5  3  3 

Total  number  of  ends  =  50.  50  4-  1.533  =  32.615,  average  num¬ 

ber.  Ans. 


§11 


YARN  CALCULATIONS,  GENERAL 


35 


56.  If  the  yarn  is  of  different  material,  such  as  cotton 
and  worsted,  then  it  is  necessary  first  to  place  the  different 
counts  in  the  same  system  before  applying  the  rule  for 
finding  the  average  number. 


Example. — There  are  placed  on  a  beam  2,000  ends  of  40s  cotton 
and  450  ends  of  45s  worsted;  what  are  the  average  counts  in  the 
cotton  system? 


Solution. — First  find  the  equivalent  cotton  counts  of  45s  worsted. 

45  X  560 

— r —  =  30s  cotton 
o4(J 

This  example  then  resolves  itself  into  finding  the  average  counts 
of  2,000  ends  of  40s  and  450  ends  of  30s. 

2  0  0  0  -=-  40  =  50 
450-T-30  =  15 
2450  65 

2,450  -5-  65  =  37.69s,  average  counts.  Ans. 


FANCY  WARPS 

57.  When  more  than  one  color  of  yarn  is  placed  on  the 
same  beam,  it  frequently  becomes  necessary  to  find  the  total 
number  of  ends  of  each  color  and  the  weight  of  each  par¬ 
ticular  yarn. 

In  order  fully  to  understand  the  explanations  given  in  this 
connection  it  will  be  necessary  first  to  consider  a  few  terms 
that  will  frequently  be  met  with.  The  yarn  that  is  placed  on 
the  loom  beam  is  known  as  the  warp,  or  warp  yarn.  It  is 
this  yarn  that  forms  the  threads  running  lengthwise  in  the 
cloth  and  is  thus  distinguished  from  the  yarn  running  across 
.the  cloth,  which  is  known  as  the  fillingr-  In  case  the  warp 
yarn  is  composed  of  different  colors  or  different  counts,  the 
order  in  which  the  different  counts  or  colors  are  placed  on 
the  beam  is  known  as  the  pattern  of  tlie  warp.  Thus,  if 
the  warp  is  arranged  4  ends  of  black,  4  ends  of  white,  4  ends 
of  black,  4  ends  of  white,  and  so  on  across  the  cloth,  the 
warp  pattern  is  said  to  be  4  black,  4  white. 

58.  To  find  the  number  of  ends  of  each  color  of  yarn  on 
a  beam  when  the  warp  pattern  and  total  number  of  ends 
are  given: 


36 


YARN  CALCULATIONS,  GENERAL 


§11 


Rule. — As  the  number  of  ends  in  one  pattern  is  to  the  number 
of  ends  of  any  one  color  in  the  pattern ,  so  is  the  total  number  of 
ends  in  the  warp  to  the  total  manber  of  ends  of  that  color. 

Example. — The  yarn  on  a  beam  is  arranged  16  ends  black,  8  ends 
white,  16  ends  black,  8  ends  gray;  how  many  ends  of  each  color  are 
there  if  there  are  2,400  ends  on  the  beam? 

Solution. —  1  6  ends  black 
8  ends  white 
1  6  ends  black 
8  ends  gray 

4  8  =  total  number  of  ends  in  one  pattern. 

There  are  32  ends  of  black  in  one  pattern. 

Therefore,  48  :  32  =  2,400  :  ;tr 


32  X  2,400  ,  AAA  ^  t  U1  , 

x  =  - -r^ - =  1,600  ends  of  black. 

4o 

There  are  8  ends  of  white  in  one  pattern. 
Therefore,  48  :  8  =  2,400  :  x 

8  X  2,400 


Ans. 


x  — 


48 


=  400  ends  of  wThite.  Ans. 


There  are  8  ends  of  gray  in  one  pattern. 
Therefore,  48  :  8  =  2,400  :  x 

8  X  2,400 


x  — 


48 


400  ends  of  gray.  Ans. 


Explanation. — The  total  number  of  ends  in  one  pattern 
must  always  bear  the  same  relation  to  the  number  of  ends  of 
any  one  color  in  the  pattern  that  the  total  number  of  ends  in 
the  warp  bears  to  the  total  number  of  ends  of  that  color. 

59.  If  it  is  desired  to  find  the  weight  of  the  ends  of  each 
color,  after  having  obtained  the  total  number  of  ends  of  each 
color,  apply  the  rule  for  finding  weight  when  length,  counts, 
and  number  of  ends  are  given. 


EXAMPLES  FOR  PRACTICE 

1.  What  is  the  weight  of  a  cotton  warp  200  yards  long  that  con¬ 
tains  2,400  ends  of  60s?  Ans.  9.52  lb. 

2.  A  cotton  warp  is  arranged  3  ends  black,  1  white,  7  black, 
1  white,  counts  24s,  154  yards  long,  2,400  ends;  find  the  weight  of 
each  color  and  the  total  weight  of  the  warp. 

f  15.27  lb.,  black 
Ans.<  3.05  lb.,  white 

1 18.32  lb.,  total  weight 


§11 


YARN  CALCULATIONS,  GENERAL 


37 


3.  A  worsted  warp  780  yards  long  contains  2,480  ends  of  20s  yarn; 

what  is  the  weight  of  the  yarn?  Ans.  172.714  lb. 

4.  What  are  the  counts  of  warp  yarn  in  a  cotton  warp  200  yards 
long  that  contains  2,400  ends  and  weighs  15  pounds?  Ans.  38.09s 

5.  A  cotton  warp  is  120  yards  long,  contains  4,480  ends,  and 

weighs  40  pounds;  what  are  the  counts?  Ans.  16s 

6.  A  worsted  warp  and  beam  weigh  170  pounds,  the  beam  alone 

weighing  70  pounds;  if  the  warp  is  350  yards  long  and  contains 
2,240  ends,  what  is  the  size  of  the  yarn?  Ans.  14s 

7.  40  pounds  of  25s  cotton  is  made  into  a  warp  containing  875  ends; 

what  is  the  length?  Ans.  960  yd. 

8.  A  woolen  warp  contains  1,648  ends  of  4j-run  yarn  and  weighs 

103  pounds;  how  long  is  the  warp?  Ans.  450  yd. 

9.  A  woolen  warp  is  320  yards  long  and  weighs  124  pounds,  exclu¬ 

sive  of  the  beam;  if  the  warp  is  composed  of  2^-run  yarn,  how  many 
ends  does  it  contain?  Ans.  1,550  ends 

10.  How  many  ends  are  there  in  a  cotton  warp  1,400  yards  long 
weighing  200  pounds,  the  counts  of  the  yarn  being  20s? 

Ans.  2,400  ends 

11.  What  are  the  average  counts  in  a  cotton  warp  arranged  24  ends 

of  40s  and  1  of  8s?  Ans.  34.48s 

12.  A  worsted  warp  contains  1,620  ends  of  45s  yarn  and  840  ends 

of  20s  yarn;  find  the  average  counts  of  the  warp.  Ans.  31.53s 

13.  A  woolen  warp  contains  900  ends  of  4^-run  yarn  and  450  ends 

of  2i-run;  find  the  average  size  of  the  yarn.  Ans.  3f-run 

14.  A  cotton  warp  is  arranged  40  ends  of  20s  and  20  ends  of  10s; 

what  are  the  average  counts?  Ans.'  15s 


38 


YARN  CALCULATIONS,  GENERAL 


§11 


METRIC  SYSTEM  OF  NUMBERING 

YARNS 

60.  In  recent  years  there  has  been  from  time  to  time 
considerable  agitation  along  the  lines  of  adopting  one  system 
of  numbering  all  classes  of  yarns  used  in  textile  manufac¬ 
turing.  The  objects  of  these  movements  are:  first,  to  bring 
about  the  unification  of  the  method  of  stating  the  degree  of 
fineness  of  the  yarns  for  all  varieties  of  fibers  used  in  the 
textile  industry  in  the  whole  world;  second,  to  adopt  a 
method  that  would  be  practical  in  every  way. 

The  advantages  of  such  a  system  as  this  would  be  many. 
The  chief  objection  to  it  is  that,  from  long  usage,  the  methods 
at  present  adopted  are  too  well  developed  for  a  single  cor¬ 
poration  or  a  single  country  to  take  on  itself  such  a  reform, 
without  being  assured  that  its  neighbors  and  competitors 
will  simultaneously  do  the  same  thing,  as  a  mill  instituting 
such  a  method  would  be  at  a  disadvantage  as  compared  with 
its  competitors. 

The  method  usually  set  forth  is  that  of  numbering  all 
classes  of  yarns  by  what  is  known  as  the  metric  system, 
in  which  1  meter  of  No.  1  yarn  weighs  1  gram,  the  meter 
being  the  unit  of  length  in  the  metric  system  and  the  gram 
the  unit  of  weight.  The  only  parts  of  the  metric  system 
that  it  is  necessary  to  understand,  in  order  to  convert  one 
system  into  the  other,  are  the  equivalents  of  the  meter  and 
the  gram,  which  are  as  follows: 

1  yard  =  .914  meter,  1  pound  =  453.59  grams 

61.  To  find  the  number  of  yarn  in  any  present  standard 
system  that  corresponds  to  the  number  of  yarn  in  the  metric 
standard  system: 

Rule. — Multiply  the  counts,  given  in  the  metric  system ,  by 
453.59  ( grams  in  1  lb.)  and  divide  by  the  standard  number  of 


§11 


YARN  CALCULATIONS,  GENERAL 


39 


yards  to  the  pound  in  the  present  system  multiplied  by  .914 
( meter  in  1  yard). 

Example. — A  cotton  yarn  numbered  according  to  the  metric  system 
is  marked  40s.  Find  the  counts  in  the  present  system. 

40  X  453.59 

Solution.—  840  X  914  =  23‘631s-  Ans- 


62.  To  find  the  number  of  yarn  in  the  metric  standard 
system  that  corresponds  to  the  number  of  yarn  in  any  present 
standard  system: 


Rule. — Multiply  the  counts ,  given  in  the  present  system,  by 
the  present  standard  number  of  yards  to  the  pound  and  by  .914 
( meter  in  1  yd.)  and  divide  by  453.59  {grams  in  1  pound). 


Example. — A  worsted  yarn  numbered  according  to  the  present 
system  is  marked  46s.  Find  the  counts  in  the  metric  system. 


46  X  560  X  .914 


51.907s.  Ans. 


Solution. — 


453.59 


CLOTH  CALCULATIONS, 

COTTON 


CALCULATIONS  NECESSARY  FOR  CLOTH 

PRODUCTION 


INTRODUCTION 

1.  Importance  of  Cloth  Calculations. — The  calcula¬ 
tions  that  come  under  the  head  of  cloth  are  by  no  means  the 
least  essential  of  the  many  dealing  with  cotton-mill  processes. 
Important  alike  to  the  designer,  superintendent,  and  treas¬ 
urer,  they  play  a  leading  part  in  the  manufacture  of  the  raw 
stock  into  the  finished  product.  Not  only  is  it  necessary  to 
ascertain  the  many  details  regarding  the  structure  of  a  piece 
of  cloth  before  it  can  be  made,  but  it  must  also  be  learned 
whether  it  is  possible  for  the  mill  to  make  such  a  cloth. 

2.  Definitions. — After  the  warp  yarn  has  been  wound 
on  the  loom  beam,  the  separate  ends  are  drawn  through  the 
harnesses  and  afterwards  through  the  reed.  The  warp  is 
then  ready  to  be  placed  in  the  loom.  The  harnesses  are 
attached  to  mechanisms  that  raise  and  lower  them;  and, 
since  some  of  the  harnesses  are  up  while  others  are  down, 
a  division  of  the  warp  yarn  must  necessarily  take  place. 
It  is  through  the  space  formed  by  this  division  that  the 
filling  passes  and,  by  this  manner  of  interlacing,  the  cloth 
is  formed. 

In  the  language  of  the  mill,  the  threads  of  a  cloth  that  run 
lengthwise  of  the  piece,  or  the  warp,  are  always  spoken  of  as 

For  notice  oi  copyright,  see  page  immediately  following  the  title  page 

2  12 


2 


CLOTH  CALCULATIONS,  COTTON 


§12 


the  ends,  while  those  that  run  from  side  to  side  are  known 
as  the  picks.  A  cloth  is  said  to  be  so  many  sley,  which 
means  that  it  contains  so  many  ends  per  inch.  It  is  also 
spoken  of  as  being  such  a  pick  cloth,  by  which  it  is  meant 
that  the  cloth  has  so  many  picks  per  inch.  Thus,  regular 
print  cloth  is  said  to  be  64-sley  and  64-pick,  which  means 
that  the  cloth  contains  64  ends  and  64  picks  per  inch;  this  is 
known  as  the  counts  of  the  cloth.  When  cloth  contains  the 
same  number  of  ends  per  inch  as  picks  it  is  spoken  of  as 
being  so  many  square.  Thus,  the  print  cloth  just  referred 
to  is  known  as  64  square. 

When  specifying  the  counts  of  a  cloth  in  writing,  the 
number  of  ends  per  inch  is  always  placed  first  and  is  fol¬ 
lowed  by  the  multiplication  sign  after  which  the  number 
of  picks  per  inch  is  placed.  Thus,  if  a  cloth  contains  80  ends 
and  60  picks  per  inch,  it  is  written  80  X  60  and,  in  speaking 
of  the  counts  of  this  cloth,  it  is  said  to  be  eighty  by  sixty. 

In  speaking  of  the  weight  of  cotton  cloth,  the  number 
of  yards  in  a  pound  is  considered  and  the  cloth  is  said  to 
be  a  so  many  yard  cloth.  Thus,  ordinary  print  cloth  is 
spoken  of  as  being  a  7-yard  cloth ,  which  means  that  it  takes 
7  yards  of  the  cloth  to  weigh  1  pound.  This  method  differs 
very  materially  from  that  in  practice  in  the  woolen  and 
worsted  trades,  where  a  cloth  is  said  to  be  a  so  many 
ounce  cloth;  that  is,  if  a  piece  of  cloth  weighs  12  ounces 
to  the  yard  it  is  said  to  be  a  12-otince  cloth.  The  weight 
of  heavy  cotton  goods,  such  as  duck,  is  indicated  by  the 
weight  of  a  square  yard;  that  is,  a  piece  of  duck  weighing 
7  ounces  to  the  square  yard,  is  spoken  of  as  7 -ounce  duck. 

The  other  specifications  necessary  in  connection  with  a 
piece  of  cloth  are  the  width,  the  counts  of  the  warp  yarn, 
and  the  counts  of  the  filling.  In  giving  these  specifica¬ 
tions  they  are  shown  as  follows:  48  X  52  —  36"  —  4.15  yards 
—  18s  warp  —  22s  filling.  The  counts  of  the  warp  and  filling 
are  sometimes  written  in  the  following  form:  18s/22s.  These 
specifications  show  that  the  cloth  is  48-sley,  52-pick,  36  inches 
wide,  4.15  yards  to  the  pound,  the  warp  being  18s,  and  the 
filling  22s. 


12 


CLOTH  CALCULATIONS,  COTTON 


3 


■8  ,  . . 

«  - . .  '  ' 

K| 

1^©»«*5^»<JI©^9004© 


HARNESS  CALCULATIONS 

3.  The  harnesses  consist  of  small  wires,  or  in  many 
cases  threads,  known  as  lieddles,  near  the  center  of  which 
eyes  are  formed,  through  which  the  warp  ends  are  drawn. 
Whenever  a  new  warp  is  drawn  in,  it  becomes  necessary  to 
find  the  number  of  heddles  that  must  be  placed  on  each  har¬ 
ness,  in  order  that  there  may  be  sufficient  heddles  for  all  the 
warp  ends  on  the  beam.  In  order  to  perform  such  a  calcula¬ 
tion  one  must  know  the  manner  of  drawing  in  the  ends, 
which  is  learned  by 
consulting  the  har¬ 
ness  draft,  which 
shows  through  which 
harness  each  end  in 
one  repeat  of  the 
draft  is  drawn. 

Fig.  1  shows  a 
harness  draft,  since 
it  indicates  through 
which  harness  the  separate  ends  are  drawn.  Each  figure 
indicates  through  which  harness  one  particular  end  is  drawn; 
thus,  the  draft  in  Fig.  1  shows  that  the  first  end  is  drawn 
through  the  first  harness;  the  second  end  through  the  second 
harness;  the  third  end  through  the  first  harness;  and  so  on 
through  the  10  ends  that  constitute  one  repeat  of  the  draft; 
that  is,  the  other  ends  in  the  warp  are  drawn  in  a  similar 
manner. 


4th  Harness 
3rd  , , 
2nd  ,, 

1  8t  9  « 


4 

3 

3 

2 

2 

2 

2 

1 

1 

1 

Fig.  1 


4.  To  find  the  necessary  number  of  heddles  on  any 
harness: 


Rule. — Find  the  number  of  repeats  of  the  harness  draft 
in  the  warp  by  dividing  the  total  number  of  ends  in  the  warp 
by  the  number  of  ends  in  one  repeat.  Multiply  the  restilt  by 
the  number  of  heddles  required  on  any  harness  for  one  repeat. 
The  result  will  be  the  total  member  of  heddles  required  on 
that  harness. 


4 


CLOTH  CALCULATIONS,  COTTON 


12 


Example. — If  a  warp  contains  2,400  ends  and  is  drawn  in  according 
to  the  draft  shown  in  Fig.  1,  how  many  heddles  should  be  placed 
on  each  harness? 

Solution. —  2,400  -s-  10  =  240  repeats  of  the  pattern. 

240  X  3  =  720  heddles  on  first  harness 

240  X  4  =  960  heddles  on  second  harness 

240  X  2  =  480  heddles  on  third  harness 

240  X  1  =  240  heddles  on  fourth  harness.  Ans. 

Explanation. — Fig.  1  shows  that  it  takes  10  ends  to  make 
one  repeat  of  the  draft;  then,  since  there  are  2,400  ends  in  the 
warp  this  draft  must  be  repeated  2,400  -f-  10  =  240  times  in 
order  to  use  all  the  warp  ends;  that  is,  the  person  drawing  in 
the  warp  ends  will  draw  the  first  10  ends  after  the  manner 
indicated  in  the  draft  and  will  then  repeat  this  operation 
with  the  next  10  ends  and  so  on  until  the  2,400  ends  have 
all  been  drawn  in.  This  will  necessitate  240  repetitions 
of  the  draft.  Now,  by  counting  across  the  first  harness,  as 
shown  in  the  draft,  it  will  be  found  that  in  one  repeat  (the 
draft  shows  one  repeat)  3  ends  are  drawn  on  this  harness. 
In  other  words,  3  heddles  must  be  placed  on  the  first  harness 
in  order  to  accommodate  the  ends  for  only  one  repeat;  and 
since  there  are  240  repeats  of  the  draft,  then  there  must  be 
240  X  3  =  720  heddles  on  the  first  harness.  Counting  across 
the  second  harness,  as  shown  in  the  draft,  it  is  seen  that  in 
one  repeat  4  ends  are  drawn  on  this  harness,  thus  necessita¬ 
ting  4  heddles;  240  repeats  will  therefore  require  240  X  4 
=  960  heddles.  The  number  of  heddles  required  on  the 
other  harnesses  is  obtained  in  the  same  manner.  As  a  rule 
a  few  extra  heddles  are  added  to  each  harness  in  order  to 
meet  all  additional  requirements. 


REEDS 

5.  The  reed,  through  which  the  ends  are  drawn  after 
being  drawn  through  the  harnesses,  plays  a  very  important 
part,  not  only  in  the  weaving,  but  also  in  all  calculations 
connected  with  cloth.  Reeds  are  made  of  thin,  flat  pieces  of 
steel  wire  set  into  top  and  bottom  pieces  known  as  ribs. 


§12 


CLOTH  CALCULATIONS,  COTTON 


5 


The  space  between  two  adjoining  wires  in  the  reed  is 
known  as  a  dent,  and  it  is  the  number  of  these  dents  that 
the  reed  contains  in  an  inch  that  determines  the  counts  of  the 
reed.  Thus,  for  example,  if  a  certain  reed  has  40  dents  per 
inch,  it  is  known  as  a  40s  reed.  In  many  cases,  however, 
reeds  are  numbered  by  giving  the  number  of  dents  in  a 
certain  number  of  inches.  For  example,  a  reed  may  be  num¬ 
bered  1,200-30,  which  indicates  that  it  contains  1,200  dents 
in  30  inches.  It  will  be  seen  that  in  both  of  these  cases  the 
counts  of  the  reed  are  the  same. 

Reeds  are  also  sometimes  spoken  of  as  being  such  a  sley; 
thus,  a  reed  may  be  said  to  be  a  64-sley,  which  means  that, 
with  the  ends  of  a  warp  drawn  in  two  per  dent,  the  cloth  will 
contain  64  ends  per  inch.  This  does  not  indicate  that  there 
are  32  dents  per  inch  in  the  reed,  since,  on  account  of  the 
contraction  that  takes  place  during  weaving,  the  yarn  at 
the  reed  is  slightly  wider  than  it  is  after  it  becomes  a  part 
of  the  cloth  and,  for  this  reason,  the  number  of  dents  per  inch 
is  slightly  less. 

Reeds  as  sent  out  by  the  manufacturers  are  always  marked 
by  one  of  the  methods  indicated  above;  that  is,  either  accord¬ 
ing  to  the  number  of  dents  per  inch  or  the  number  of  dents 
in  so  many  inches.  However,  reeds  are  sold  by  the  bier . 
The  bier,  as  applied  to  reeds,  means  20  dents;  consequently, 
when  the  price  per  reed  is  quoted  at  so  much  per  bier,  it 
means  so  much  for  every  20  dents. 


CALCULATIONS  FOR  WARP  YARNS 


FINDING  THE  NUMBER  OF  ENDS  IN  A  WARP 

6.  The  first  calculation  that  is  necessary  when  dealing 
with  cloth  and  one  that  will  be  found  to  be  as  simple  as  any 
is  that  of  learning  the  total  number  of  ends  in  a  warp  when 
the  width  of  the  cloth  and  the  ends  per  inch,  or  the  sley  of 
the  cloth,  are  given.  Suppose,  for  example,  that  a  cloth  is 
30  inches  wide  and  64-sley.  Since  there  are  64  ends  in  1  inch, 


6  CLOTH  CALCULATIONS,  COTTON  §12 

it  naturally  follows  that  in  30  inches  there  are  30  X  64 
=  1,920  ends. 

One  point,  however,  should  be  noted  in  this  connection — 
at  the  sides  of  all  cloths  additional  ends  are  placed  in  order 
to  strengthen  the  cloth.  These  ends  are  known  as  the  sel¬ 
vage  ends,  and  in  making  calculations  to  determine  the 
number  of  ends  in  a  piece  of  cloth  it  is  always  necessary  to 
consider  these.  They  are  generally  ends  like  those  of  the 
body  of  the  warp,  where  such  ends  are  all  alike,  or  like 
those  forming  the  plain  portion  of  a  fancy  cloth  containing 
several  varieties  or  counts  of  warp  yarn,  but  are  usually 
reeded  with  twice  as  many  ends  per  dent  as  similar  ends  in 
the  body  of  the  warp;  thus,  if  a  warp  is  drawn  in  two  per 
dent,  for  about  i  inch  in  width  at  each  side  the  ends  will 
be  drawn  in  four  per  dent.  Generally  speaking,  from  16  to 
24  additional  ends  on  each  side  of  the  cloth  will  be  found  to 
be  sufficient  to  allow  for  the  selvages. 

7.  To  find  the  total  ends  in  a  cloth  when  the  width  and 
sley  are  given. 

Rule. — Multiply  the  sley  by  the  width  a?id  to  the  result  thus 
obtained  add  a  certain  number  of  ends  for  selvages. 

Example.  —  Find  the  total  number  of  ends  in  a  cloth  30  inches  wide 
and  containing  64  ends  per  inch. 

Solution.—  30  X  64  =  1,920.  Considering  that  there  are  24  extra 
ends  on  each  side  for  selvages,  making  48  additional  ends  for  selvages, 
1,920  +  48  =  1,968  ends  in  the  cloth.  Ans. 

8.  Suppose  that  a  small  piece  of  cloth  containing  a  warp 
pattern  is  all  that  a  person  has  to  figure  from,  and  that  it  is 
desired  to  learn  how  many  ends  of  each  kind  the  warp  must 
contain;  it  is  first  necessary  to  determine  the  width  that  the 
cloth  is  to  be.  This  will,  of  course,  be  given  by  the  person 
ordering  the  cloth.  Then  determine  the  number  of  ends  of 
each  kind  in  one  repeat  of  the  pattern.  To  illustrate,  sup¬ 
pose  that  a  small  sample  is  found  to  contain  for  a  warp 
pattern  48  ends  cotton,  12  ends  mercerized,  and  that  it  is 
decided  to  make  the  cloth  30  inches  wide  inside  the  selvages. 
The  cloth  is  all  produced  from  cotton,  but  a  portion  of  the 


12 


CLOTH  CALCULATIONS,  COTTON 


7 


yarn  will  have  passed  through  a  mercerizing  process  to  give 
it  a  glossy  appearance. 

A  rule  has  been  given  for  finding  the  number  of  ends  of 
each  count  or  color  on  a  beam  when  different  counts  or 
colors  are  used.  When,  however,  these  items  are  to  be 
figured  from  a  piece  of  cloth  it  is  better  to  employ  slightly 
different  methods. 

Measure  the  space  occupied  by  one  pattern.  To  obtain 
accurate  results  it  is  best  to  use  a  pair  of  dividers  and  a 
small  steel  rule.  Suppose  the  pattern  to  measure  i  inch. 
Next  find  the  number  of  patterns  that  the  cloth  contains,  by 
dividing  the  entire  width  by  the  width  of  one  pattern,  which 
gives  30  -r-  i  =  60  patterns  in  the  cloth.  Multiplying  the 
number  of  patterns  in  the  cloth  by  the  number  of  ends  of 
each  kind  in  one  pattern  gives  the  total  number  of  each  kind 
df  ends  in  the  cloth.  Thus, 

60  X  48  =  2,880  ends  of  cotton 
60  X  12  =  720  ends  of  mercerized 

3,600  ends  in  the  cloth  +  48  ends  for  selvages 
equals  3,648,  the  total  number  of  ends  required  for  the  cloth. 

Note. — In  all  calculations  of  this  nature  it  is  important  to  remember 
that  the  space  occupied  by  the  patterns  does  not  include  the  space 
occupied  by  the  selvage  ends.  Selvage  ends  should  be  calculated 
separately  from  those  forming  the  body  of  the  warp. 

9.  To  find  the  total  number  of  ends  of  each  kind  in  the 
body  of  the  cloth  when  different  counts,  colors,  or  materials 
are  used: 

Rule. — Find  the  number  of  patterns  in  the  cloth  by  dividing 
the  width,  exclusive  of  selvages ,  by  the  width  of  one  pattern , 
and  multiply  the  result  by  the  number  of  ends  of  each  kind  in 
one  pattern.  _ 

FINDING  NUMBER  OF  HANKS  OF  WARP  YARN 

10.  Contraction. — If  it  should  be  desired  to  find  the 
amount  of  warp  yarn  required  for  a  cut,  say  50  yards  long, 
of  any  cloth,  it  would  naturally  appear  that  all  that  would 
be  necessary  in  order  to  obtain  the  total  length  of  warp  yarn 


8 


CLOTH  CALCULATIONS,  COTTON 


12 


would  be  to  multiply  the  number  of  ends  in  the  cloth  by  the 
length  of  the  cut,  after  which  the  weight  or  number  of  hanks 
could  be  figured  by  the  rules  previously  given. 

In  this  connection,  however,  it  is  very  essential  to  take 
into  account  the  contraction  of  the  warp,  which  takes  place 
during  weaving.  This  contraction  affects  both  the  length 
and  width  of  the  cloth,  but  since  only  the  warp  yarns  are 
being  dealt  with  it  is  only  necessary  at  present  to  consider 
the  contraction  in  the  length. 

Since,  in  the  interlacing  of  the  filling  with  the  warp,  the 
two  series  of  yarns  necessarily  bend  around  each  other  to  a 
certain  extent,  it  naturally  follows  that  a  piece  of  cloth  will 
not  be  quite  as  long  as  the  warp  yarns  from  which  it  is 
made.  For  example,  if  the  warp  ends  on  a  beam  are 
100  yards  long,  the  cloth  woven  from  this  length  of  yarn 
will  be  slightly  shorter.  This  difference  between  the  length 
of  the  warp  yarn  and  the  cloth  made  from  it  is  known  as 
the  contraction. 

The  main  factors  that  will  tend  to  either  increase  or 
decrease  the  amount  of  contraction  that  takes  place  are 
considered  below. 

The  tension  on  the  .yarn  during  weaving  will  affect  the 
contraction  to  a  slight  extent.  The  comparative  counts  of 
the  warp  and  filling  also  have  an  influence,  since,  if  the 
warp  is  very  much  coarser  than  the  filling,  the  filling  will 
do  most  of  the  bending,  while  the  warp  yarn  will  lie  in  a 
comparatively  straight  line.  The  class  of  weave,  or  in  other 
words  the  manner  of  interlacing  the  warp  and  filling,  will 
also  have  considerable  influence  on  the  amount  of  contrac¬ 
tion,  since  the  warp  yarn  will  not  contract  so  much  in  a 
weave  where  it  interlaces  with  the  filling  only  once  in  five 
picks  as  it  will  in  a  weave  where  it  interlaces  at  every  pick. 
Moreover,  weaves  in  which  the  warp  yarns  are  drawn  entirely 
out  of  a  straight  line,  such  as  lenos,  will  contract  the  warp 
yarn  much  more  than  will  weaves  in  which  the  warp  yarns 
lie  in  a  comparatively  straight  line. 

The  following  rule  will  aid  in  determining  what  percentage 
should  be  allowed  for  contraction  of  warp  yarn. 


§12  CLOTH  CALCULATIONS,  COTTON  9 

11.  To  find  the  percentage  of  warp  contraction  during 
weaving: 

Rule. — Multiply  the  number  of  picks  per  inch  by  3  and 
divide  by  the  counts  of  the  filling.  The  result  will  be  the  per¬ 
centage  to  allow  for  contraction. 

Example. — The  number  of  picks  per  inch  in  a  certain  cloth  is  60, 
the  counts  of  the  filling  are  36s;  what  will  be  the  length  of  the  cloth 
made  from  100  yards  of  warp  yarn? 

0  60  X  3  _  ,  .  ,  .  ,  „ 

Solution. —  — — —  =  5,  which  is  the  percentage  to  allow  for  con¬ 

traction.  5  per  cent,  of  100  =  5.  100  yd.  —  5  yd.  =  95  yd.  of  cloth  from 
100  yd.  of  warp. 

This  rule,  when  taking  into  consideration  the  points  pre¬ 
viously  mentioned,  is  comparatively  accurate  for  counts  of 
filling  from  25s  to  50s  and  for  picks  from  40  to  80  per  inch  and 
will  serve  as  a  basis  when  finding  the  contraction  of  any  warp. 

By  varying  the  constant  3  to  suit  special  circumstances  the 
student  can  formulate  rules  to  suit  requirements;  or  if  the 
usual  rate  of  contraction  in  a  certain  mill  on  certain  goods  is 
found,  it  will  not  be  difficult  to  form  a  good  idea  of  the  con¬ 
traction  in  other  cloths  when  they  are  met  with.  However, 
when  dealing  with  lenos  and  like  weaves  in  which  the  con¬ 
traction  is  very  great,  the  best  plan  to  follow  is  to  measure 
the  space  that  an  end  occupies  when  in  the  cloth,  after  which 
take  out  the  end  and  measure  its  length  when  pulled  straight. 
By  this  means  it  is  a  comparatively  easy  matter  to  learn  the 
amount  of  contraction. 

12.  To  find  the  hanks  of  warp  yarn  contained  in  a  cut 
of  cloth  of  any  length: 

Rule. — Multiply  the  number  of  ends  in  the  cloth  by  the  length 
of  the  warp  yarn  in  the  cut  before  weaving  and  divide  by  the 
standard  number  of  yards. 

Example. — From  an  example  previously  given  it  was  learned  that 
a  cloth  30  inches  wide  having  64  ends  per  inch  contained,  with  the 
ends  for  selvages,  a  total  of  1,968  ends;  if  this  cloth  is  woven  50  yards 
long  from  52  yards  of  warp,  how  many  hanks  of  warp  does  it  contain  ? 

Solution. —  1,968  X  52  =  102,336  yd. 

102,336  -4-  840  =  121.828  hanks.  Ans. 


10 


CLOTH  CALCULATIONS,  COTTON 


§12 


Explanation. — There  are  1,968  ends  in  the  warp;  if  each 
end  were  1  yard  long,  there  would  be  1,968  yards,  but  since 
each  end  is  52  yards  long  (the  length  before  contraction 
must  be  taken)  there  must  be  52  X  1,968  yards  =  102,336 
yards  of  warp.  As  it  always  takes  840  yards  to  make  a 
hank  of  cotton  yarn  no  matter  what  the  counts  may  be,  in 
order  to  learn  the  number  of  hanks  in  this  amount  of  yarn, 
the  total  number  of  yards  must  be  divided  by  the  number 
of  yards  in  1  hank.  _ 

CALCULATIONS  FOR  FILLING  YARN 


WIDTH  AT  REED 

13.  When  figuring  to  ascertain  the  amount  of  filling  that 
a  cut  of  cloth  contains,  practically  the  same  particulars  are 
considered  as  in  the  contraction  of  the  warp.  Thus,  if  a 
cloth  is  30  inches  wide,  the  space  that  the  warp  yarn  occupies 
in  the  reed,  or,  as  it  is  known,  the  width  at  the  reed,  will 
be  slightly  in  excess  of  this  width.  Consequently,  to  find  the 
exact  length  of  1  pick  of  filling,  it  is  necessary  to  take  this 
width  at  the  reed  and  not  the  width  of  the  cloth.  In  order 
to  find  the  width  at  the  reed,  however,  it  is  first  necessary  to 
ascertain  the  number  of  dents  per  inch  in  the  reed  or,  in 
other  words,  the  counts  of  the  reed. 

14.  To  find  the  dents  per  inch  in  a  reed  to  produce  a 
cloth  of  a  given  sley: 

Rule. — Subtract  1  from  the  sley  of  the  cloth ,  divide  the 
result  by  the  number  of  ends  per  dent ,  arid  multiply  the  resjilt 
thus  obtained  by  .95. 

Example  1.— A  cloth  is  64-sley,  reeded  two  in  a  dent;  what  counts 
of  reed  will  be  necessary  to  give  this  sley? 

Solution. —  64  —  1  =  63 

63  2  =  31.5 

31.5  X  .95  =  29.92,  or  30  dents  per  in.  Ans. 

Explanation. — By  always  subtracting  1  from  the  sley  of 
the  cloth  a  sliding  scale  is  obtained,  which  to  a  certain 


12 


CLOTH  CALCULATIONS,  COTTON 


11 


extent  offsets  the  difference  in  the  contraction  of  different 
counts  of  yarn.  Thus,  if  the  sley  is  50  and  1  is  subtracted, 
2  per  cent,  is  deducted,  whereas  if  the  sley  is  100  apd  1  is 
subtracted,  only  1  per  cent,  is  deducted.  Since  there  are 
2  ends  per  dent,  the  sley  must  be  divided  by  this  number  in 
order  to  obtain  the  dents  that  are  occupied  by  1  inch  of  warp 
yarn  as  measured  in  the  cloth.  5  per  cent,  is  a  safe  estimate 
of  the  contraction  that  takes  place  when  running'  medium 
counts  of  yarns;  therefore,  the  result  obtained  by  dividing 
by  2  is  multiplied  by  .95  in  order  to  obtain  the  dents  per  inch. 

In  many  cases  the  warp  ends  are  drawn  more  than  two 
per  dent  throughout  the  reed.  Under  such  circumstances  it 
is  always  necessary  to  divide  the  result  obtained  by  sub¬ 
tracting  1  from  the  sley  by  whatever  number  of  ends  there 
are  to  each  dent. 

Example  2. — A  cloth  is  64-sley,  reeded  four  per  dent;  what  counts 
of  reed  will  be  necessary  to  give  this  sley? 

Solution.—  64  —  1  =  63 

63  -r-  4  =  15.75 

15.75  X  .95  =  14.96,  or  15  dents  per  in.  Ans. 

15.  To  find  the  width  occupied  by  the  warp  yarn  in  the 
reed  exclusive  of  selvages: 

Rule. — Subtract  the  number  of  selvage  ends  from  the  total 
number  of  ends  in  the  warp.  Divide  this  result  by  the  number 
of  ends  per  de?it  and  divide  the  result  thus  obtained  by  the  num¬ 
ber  of  dents  per  inch  in  the  reed. 

Example. — A  cloth  contains  2,048  ends,  including  48  selvage  ends, 
is  80-sley,  and  reeded  two  per  dent;  what  are  the  counts  of  the  reed 
and  what  is  the  width  in  the  reed? 

Solution. — 

80  —  1  =  79;  79  -p  2  =  39.5 

39.5  X  .95  =  37.52,  or  37.5  dents  per  in.  in  the  reed 
2,048  —  48  for  selvage  =  2,000  ends 
2,000  -r-  2  =  1,000  dents  required  for  the  warp 
1,000  -T-  37.5  =  26f  in.,  width  at  reed.  Ans. 

Explanation. — It  is  first  necessary  to  obtain  the  number 
of  dents  per  inch  in  the  reed;  this  is  found  by  following  the 


12 


CLOTH  CALCULATIONS,  COTTON 


§12 


rule  previously  given.  If  the  total  number  of  dents  occupied 
by  the  warp  yarn  were  known,  it  would  be  a  simple  matter 
to  find  the  width  at  the  reed,  since  all  that  would  be  neces¬ 
sary  would  be  to  divide  this  number  of  dents  by  the  dents  in 
1  inch.  To  find  the  number  of  dents  occupied  by  the  body 
of  the  warp,  the  number  of  selvage  ends  must  be  deducted. 
After  this,  divide  the  number  of  warp  ends  remaining  by  the 
number  of  ends  drawn  in  1  dent,  which  will  give  the  dents 
required.  _ 


FINDING  THE  HANKS  OF  FILLING 

16.  To  find  the  hanks  of  filling  contained  in  a  cloth  of 
any  length: 

Rule. — Multiply  the  width  in  the  reed,  in  inches ,  by  the 
number  of  picks  per  inch.  Multiply  this  result  by  the  length 
of  the  cloth,  in  yards,  and  divide  the  result  thus  obtained  by  the 
number  of  yards  to  the  hank. 


Example. — It  is  desired  to  learn  how  much  filling  there  will  be  in 
a  50-yard  cut  of  the  cloth  designated  in  Art.  15,  the  cloth  to  contain 
90  picks  per  inch. 


Solution. — 


26f  X  90  =  2,400 


2,400  X  50 
840 


142.85  hanks. 


Ans. 


Explanation. — Since  each  pick  of  filling  is  26f  inches 
long  and  there  are  90  picks  per  inch,  in  1  inch  of  cloth  there 
will  be  90  X  26f  =  2,400  inches  of  filling.  If  there  is  2,400 
inches  of  filling  in  1  inch  of  cloth,  there  must  be  2,400  yards 
of  filling  in  1  yard  of  cloth;  and  in  50  yards  there  will  be 
50  X  2,400  =  120,000  yards  of  filling.  Dividing  this  by  840 
to  obtain  the  hanks,  120,000  -f-  840  =  142.85  hanks  of  filling. 

It  will  be  noticed  that  in  figuring  for  filling  the  length  of 
the  cloth  is  used  and  not  the  length  of  the  warp. 


IRREGULAR  REED  DRAFTS 

17.  Whenever  the  warp  ends  are  drawn  in  an  irregular 
manner,  as  frequently  occurs,  it  becomes  necessary  to  follow 
a  method  slightly  different  from  that  just  described. 


12 


CLOTH  CALCULATIONS,  COTTON 


13 


Example. — A  piece  of  cloth  is  found  to  contain  the  following  for 
a  warp  pattern:  60  ends  cotton,  40  ends  silk,  24  ends  cotton,  and 
20  ends  silk;  the  cotton  ends  are  drawn  through  the  reed  two  per 
dent,  while  the  silk  ends  are  reeded  four  per  dent;  the  reed  used  con¬ 
tains  40  dents  per  inch  and  the  width  at  the  reed,  excluding  selvages, 
is  32  inches.  How  many  ends  of  each  does  the  cloth  contain? 

Solution. — First  find  the  number  of  dents  occupied  by  one  pattern: 

60  ends  cotton,  two  in  a  dent  =  30  dents 
40  ends  silk,  four  in  a  dent  =10  dents 
24  ends  cotton,  two  in  a  dent  =  12  dents 
20  ends  silk,  four  in  a  dent  =  5  dents 

Dents  in  one  pattern  =57  dents 

40  dents  per  inch  multiplied  by  32,  which  is  the  width  of  the  reed 
occupied  by  the  body  of  the  warp,  gives  1,280  as  the  total  number  of 
dents  occupied.  Dividing  this  by  the  number  of  dents  in  one  pattern 
will  give  the  number  of  patterns  in  the  warp. 

1,280  -4-  57  =  22  patterns  and  26  dents  extra 

There  are  84  ends  of  cotton  in  one  pattern  and  in  22  patterns  there 
will  be  22  X  84  =  1,848  ends  of  cotton. 

There  are  60  ends  of  silk  in  one  pattern;  therefore,  in  22  patterns 
there  will  be  22  X  60  =  1,320  ends  of  silk. 

There  are  still  26  dents  that  have  not  been  used.  In  all  cases  of 
this  kind,  it  is  best  to  have  the  plain  weave  come  next  to  the  selvage  if 
it  is  possible.  In  this  case  after  completing  the  22  patterns,  the  next 
ends  to  be  drawn  in  will  be  cotton,  and  it  will  be  necessary  to  have 
52  ends  for  the  26  dents.  This  gives  1,848  +  52  =  1,900  ends  of  cotton. 
Add  to  this  number  48  for  selvage  which  gives  1,900  +  48  =  1,948  as 
the  total  number  of  cotton  ends  in  the  warp.  1,948  ends  cotton  +  1,320 
ends  silk  =  3,268,  total  number  of  ends  in  the  warp.  Ans. 


FINDING  YARDS  PER  POUND 

18.  Another  calculation  that  is  often  found  necessary 
when  dealing  with  cloth  is  the  finding  of  the  number  of 
yards  per  pound  from  a  small  sample  of  cloth,  very  fre¬ 
quently  only  1  square  inch  of  cloth  being  given,  although 
of  course  more  accurate  results  can  be  obtained  when  larger 
pieces  are  weighed. 

19.  To  find  the  number  of  yards  per  pound  from  a  small 
sample  of  cloth: 


14 


CLOTH  CALCULATIONS,  COTTON 


12 


Rule. — Multiply  7 ,000  by  the  number  of  square  inches  weighed 
and  divide  the  result  thus  obtained  by  the  product  of  the  weight , 
in  grains ,  of  the  piece  weighed ,  the  width  of  the  cloth,  a?id  36 
( the  number  of  inches  in  one  yard) . 

Example. — Two  square  inches  of  cloth  is  found  to  weigh  4  grains; 
what  are  the  yards  per  pound  if  the  cloth  is  30  inches  wide? 

Solution.—  |  Q  =  3.24  yd.  per  lb.  Ans. 

Explanation. — Since  the  cloth  is  30  inches  wide,  there 
will  be  30  X  36  =  1,080  square  inches  of  cloth  in  1  yard. 
As  2  square  inches  weighs  4  grains,  1  square  inch  will  weigh 
2  grains.  Multiply  the  number  of  square  inches  in  1  yard 
by  the  weight  of  1  square  inch,  which  will  give  the  weight  of 
1  yard.  Since  there  is  7,000  grains  to  the  pound,  by  dividing 
7,000  by  the  weight  of  1  yard,  the  yards  per  pound  will  be 
obtained. 

Note. — In  calculations  of  this  kind  it  is  important  to  use  the  correct 
number  of  square  inches  weighed;  thus  a  piece  of  cloth  3  inches 
square  is  3  inches  long  and  3  inches  wide,  and  therefore  contains  3x3 
=  9  square  inches,  instead  of  3  square  inches. 


EXAMPLES  FOR  PRACTICE 

1.  What  is  the  weight,  in  ounces,  of  warp  yarn  in  1  yard  of  cloth 

30  inches  wide  at  the  reed,  drawn  in  a  reed  containing  30  dents  per 
inch  and  3  ends  per  dent,  the  counts  being  36s?  Ans.  ly  oz. 

2.  A  warp  64  inches  wide  at  the  reed  is  made  from  12s  yarn,  weighs 

45  pounds,  and  is  drawn  4  ends  per  dent  in  a  reed  containing  12  dents 
per  inch.  What  is  the  length  of  the  warp?  Ans.  147.65  yd. 

3.  A  warp  is  198  yards  long,  weighs  25  pounds,  and  is  made  from 

18s  yarn:  (a)  How  many  ends  does  it  contain?  (£)  How  many 
inches  wide  is  it  in  the  reed  if  woven  2  ends  per  dent  in  a  reed  contain¬ 
ing  25  dents  per  inch?  f  (a)  1,909  ends  (practically) 

1(d)  38.18  in. 

4.  What  weight  of  30s  cotton  filling  will  be  required  for  a  cut  of 

cloth  100  yards  long,  30  inches  wide  at  reed,  and  containing  56  picks 
per  inch?  Ans.  6f  lb. 

5.  A  square  inch  of  cloth  weighs  ly  grains;  what  will  be  the 
weight,  in  ounces,  of  a  yard  if  the  cloth  is  56  inches  wide? 

Ans.  6.914  oz. 


§12 


CLOTH  CALCULATIONS,  COTTON 


15 


6.  How  many  ends  are  required  to  weave  a  cloth  75-sley,  32  inches 

wide,  allowing  36  extra  ends  for  selvages?  Ans.  2,436  ends 

7.  How  many  ends  are  required  to  weave  a  cloth  60-sley,  30  inches 

wide,  allowing  40  extra  ends  for  selvages?  Ans.  1,840  ends 

8.  How  many  dents  per  inch  should  a  reed  contain  to  weave  a 
50-sley  cloth,  ends  drawn  two  per  dent?  Ans.  23.275  dents  per  in. 

9.  What  will  be  the  dents  per  inch  in  a  reed  to  weave  a  92-sley 
cloth,  ends  drawn  2  in  a  dent? 

Ans.  43.225  dents  per  in. 

10.  What  will  be  the  dents  per  inch  in  a  reed 
to  weave  a  108-sley  cloth,  ends  drawn  3  in  a  dent? 

Ans.  33.877  dents  per  in. 

11.  A  cut  of  cloth  50  yards  long,  and  34.5 
inches  wide  at  the  reed  is  woven  with  76  picks  per 
inch  of  60s  filling;  what  is  the  weight  of  the  filling?  Ans.  2.601  lb. 

12.  A  warp  containing  2,400  ends,  excluding  selvages,  is  drawn 
in  according  to  the  draft  in  Fig.  2;  how  many  heddles  will  be  required 
on  each  harness? 

A  f  300  heddles  on  first  and  fifth  harnesses 
ns'\600  heddles  on  second,  third,  and  fourth  harnesses 


5 

4 

4 

3 

3 

2 

2 

1 

Fig.  2 


16  CLOTH  CALCULATIONS,  COTTON 


§12 


FIGURING  PARTICULARS  FROM 
CLOTH  SAMPLES 


SLEY 

20.  In  order  that  the  methods  employed  when  finding 
the  particulars  of  manufacture  from  a  small  sample  of  cloth 
may  be  fully  understood,  a  piece  of  cloth  is  given  here  and 
the  particulars  worked  out,  illustrating  each  point  thoroughly. 

When  a  small  piece  of  cloth  is  given  from  which  to  pro¬ 
duce  a  similar  cloth,  the  particulars  that  must  be  learned  are 
the  sley ,  pick,  average  counts ,  number  of  yards  per  pound ,  and 
width  of  the  goods. 

The  first  item  necessary  to  obtain  is  the  sley  of  the  cloth. 
In  ordinary  plain  cloth  the  best  method  for  finding  the  sley 
is  to  use  a  pick  glass,  or,  in  some  cases,  to  cut  out  a  small 
piece  of  cloth,  say  1  or  2  inches  square,  pulling  out  the 
threads  one  by  one  and  counting  them  and  in  this  manner 
obtaining  the  number  of  ends  per  inch  in  the  cloth.  With 
the  sample  that  is  attached  here  it  will  be  impossible  to 
obtain  the  sley  by  either  of  these  methods  on  account  of 
the  number  of  ends  that  are  drawn  in  one  part  of  the  cloth 
differing  from  the  number  that  are  drawn  in  another  section. 

When  finding  the  sley  in  cases  of  this  kind  it  is  necessary 
to  proceed  as  follows:  Measure  the  plain  stripe  of  the  cloth; 
it  is  found  to  be  A  inch.  Then,  with  a  pick  glass,  count  the 
number  of  ends  that  are  included  in  this  space;  it  will  be 
seen  that  there  are  12.  There  being  12  ends  in  A  inch,  it  is 
possible  to  obtain  the  number  of  ends  that  would  be  included 
in  1  inch  if  the  entire  warp  were  drawn  in  in  this  manner. 
Since  there  are  12  ends  in  A  inch,  in  A*  inch  there  will  be 
i  of  12,  or  li,  and  in  tI,  or  1  inch,  there  will  be  48  X  Is  =  72 
ends.  Therefore,  if  the  entire  warp  were  drawn  in  plain, 
the  sley  of  the  cloth  would  be  72  and  the  number  of  dents 


12 


CLOTH  CALCULATIONS,  COTTON 


17 


that  would  be  occupied  in  the  cloth  per  inch  would  be  i  of 
72,  or  36,  provided  that  the  cloth  were  reeded  two  in  a  dent. 
It  must  be  understood  that  this  36  dents  per  inch  does  not 
mean  the  number  of  dents  per  inch  in  the  reed,  but  refers  to 
the  number  of  dents  per  inch  as  represented  by  the  warp  ends 
after  being  woven. 


REED  TABEE 

21.  After  obtaining  the  sley  it  is  necessary  to  make  out 
what  is  known  as  a  reed  table  in  the  following  manner: 

Since  the  space  occupied  by  the  plain  stripe  was  measured 
in  forty-eighths  of  an  inch,  the  reed  table  would  be  best 
made  out  in  forty-eighths.  By  dividing  36  (the  number  of 
dents  per  inch)  by  48,  the  number  of  dents  that  will  be 
occupied  by  ts  inch  will  be  obtained.  This  gives 
Therefore,  in  its  inch  in  the  cloth,  dent  will  be  taken  up. 
From  this  the  following  reed  table  is  obtained: 

Ts  inch  of  cloth  will  occupy  f  dent. 

A  inch  of  cloth  will  occupy  h§  dents. 

A  inch  of  cloth  will  occupy  2\  dents, 
inch  of  cloth  will  occupy  3  dents. 

A  inch  of  cloth  will  occupy  3-?  dents. 

-ra  inch  of  cloth  will  occupy  4|  dents. 

A  inch  of  cloth  will  occupy  5^  dents. 

A  inch  of  cloth  will  occupy  6  dents. 

A  inch  of  cloth  will  occupy  6f  dents. 

M  inch  of  cloth  will  occupy  7|  dents. 

From  this  table,  it  is  possible  after  measuring  any  part  of 
the  cloth  less  than  If  inch  to  tell  exactly  how  many  dents 
that  space  will  occupy;  or,  in  other  words,  it  is  possible  to 
tell  in  how  many  dents  the  number  of  ends  in  that  space  are 
drawn,  from  which  it  is  possible  to  learn  the  number  of 
ends  per  dent. 

Next  measuring  the  stripe,  which  contains  more  ends  per 
inch  than  the  plain,  it  is  found  to  be  its  inch  wide.  Referring 
to  the  reed  table,  it  is  found  that  it  inch  contains  5i  dents. 
This  would  be  called  5  dents.  Counting  the  number  of  ends 
in  the  stripe,  it  is  found  to  contain  25.  Therefore,  these 


18 


CLOTH  CALCULATIONS,  COTTON 


ends  are  drawn  in  five  per  dent.  It  has  now  been  learned  that 
the  number  of  ends  in  one  pattern  equals  25  +  12  =  37,  and 
also  that  the  number  of  dents  that  one  pattern  occupies  equals 
5  +  6  =  11  dents. 


FINDING  IIANKS  OF  WARP  YARN 

22.  It  is  next  necessary  to  learn  the  number  of  lianks 
of  warp  yarn  that  the  cloth  contains.  It  has  been  shown 
how  to  obtain  the  amount  of  warp  in  a  cloth,  and  in  this  case 
proceed  in  a  similar  manner. 

First  find  the  number  of  dents  that  the  whole  warp  occupies, 
by  multiplying  the  dents  per  inch,  or  36,  by  the  width.  In 
this  case  it  will  be  assumed  that  the  cloth  is  27  inches  wide 
inside  selvages.  Therefore,  the  following  is  obtained: 
36  X  27  =  972,  or  972  dents  occupied  by  the  cloth. 

Since  there  are  11  dents  to  one  pattern,  to  find  the  number 
of  patterns  that  this  warp  occupies,  divide  the  total  number 
of  dents  by  the  number  of  dents  to  one  pattern:  972  -f-  1 1  =  88 
patterns  and  4  dents  over. 

Since. there  are  37  ends  in  one  pattern,  in  order  to  find  the 
total  number  of  ends  in  the  warp,  multiply  the  number  of 
patterns  by  the  number  of  ends  in  each  pattern:  88  X  37 
=  3,256. 

Four  dents  remain  after  obtaining  the  88  patterns.  In  all 
cases  of  this  kind  it  is  a  common  custom  to  use  these  extra 
dents  to  draw  in  the  warp  plain,  2  ends  in  a  dent,  thus 
having  a  plain  stripe  at  each  side  of  the  cloth  in  place  of  any 
fancy.  Following  this  rule  in  this  case,  the  4  dents  extra 
drawn  two  per  dent  will  give  8  extra  ends.  Adding  these  to 
the  3,256  already  obtained  will  give  3,264  ends  in  the  warp, 
but,  as  previously  explained,  it  is  necessary  to  add  to  this 
number  the  number  of  ends  that  are  used  for  selvages.  In 
this  case  48  ends  will  be  used,  which  added  to  the  3,264  already 
obtained  gives  3,312  as  the  total  number  of  ends  in  the  warp. 

Note. — In  all  the  calculations  of  this  cloth,  it  must  be  kept  in  mind 
that  the  48  selvage  ends  are  reeded  four  per  dent,  thus  adding  12  dents 
to  the  number  of  dents  occupied  in  the  reed  by  the  body  of  the  cloth 
and  consequently  adding  ^  inch  to  the  width  previously  given  of  this 
part  of  the  woven  cloth. 


12 


CLOTH  CALCULATIONS,  COTTON 


19 


In  figuring  to  obtain  any  of  the  particulars  for  a  sample  of 
cloth  it  is  always  the  best  plan  to  take  a  number  of  yards 
greater  than  one  and  figure  with  this  as  a  basis.  Any  errors 
that  may  occur  on  account  of  the  difference  in  contraction 
and  like  subjects  will  be  reduced  by  this  means  to  a  minimum. 
In  figuring  this  cloth,  a  cut  50  yards  long  will  be  taken  as  a 
basis  for  all  calculations. 

Making  allowance  for  the  contraction  that  takes  place  dur¬ 
ing  weaving  by  considering  that  53  yards  of  warp  will 
produce  50  yards  of  cloth,  it  becomes  necessary  to  find  the 
number  of  yards  of  yarn  that  will  be  contained  in  50  yards 
of  cloth.  It  has  already  been  found  that  there  are  3,312  ends 
in  the  warp.  Multiplying  this,  then,  by  the  number  of  yards 
of  warp  yarn  will  give: 

3,312  X  53  =  175,536  yards  of  warp  in  the  cloth 
175,536  -7-  840  =  208.97  hanks  of  warp  yarn  in  the  cloth 


ALLOWANCE  FOR  SIZE 

23.  One  point  needs  to  be  noted  carefully  by  the  stu¬ 
dent — that,  before  weaving,  size  is  placed  on  the  warp  yarn, 
which  adds  somewhat  to  its  weight.  The  American  custom 
of  sizing  yarn  differs  considerably  from  that  employed  in 
Europe  and  some  other  foreign  countries,  where  size  is  largely 
added  for  the  purpose  of  weighting  the  cloth.  In  America, 
however,  the  principal  use  of  size  is  for  the  purpose  of 
strengthening  the  warp  yarn  in  order  to  enable  it  to  undergo 
the  strain  and  chafing  that  take  place  during  weaving,  and, 
since  for  this  purpose  the  amount  of  size  added  is  very  much 
smaller  than  that  used  when  sizing  for  weight,  only  the 
American  custom  will  be  considered  here. 

If  the  percentage  of  size  added  were  calculated  from  the 
weight  of  the  cloth,  the  result  would  not  be  correct,  since  the 
size  is  added  only  to  the  warp  yarn  and  not  to  the  filling. 
Therefore,  in  order  to  allow  for  this  additional  weight  of 
size,  it  is  necessary  to  employ  some  other  method. 

The  method  followed  is  to  add  the  percentage  of  size  to 
the  hanks  of  warp  yarn.  Since  hanks  of  warp  divided  by 


20 


CLOTH  CALCULATIONS,  COTTON 


§12 


the  counts  of  yarn  will  give  the  weight,  it  will  be  seen 
that  if,  instead  of  adding  the  percentage  of  size  to  the 
weight,  it  is  added  to  the  hanks  of  warp  the  same  result 
will  be  obtained.  Therefore,  after  obtaining  the  hanks  of 
warp  yarn  in  the  cloth,  add  a  certain  per  cent,  for  size. 
Generally  it  will  be  found  that  10  per  cent,  will  cover  all  cases; 
this  is  the  amount  allowed  in  this  case:  208.97  hanks  of 
warp  +  10  per  cent,  for  size  equals,  in  weight,  229.867  hanks 
of  unsized  warp  yarn. 


FINDING  HANKS  OF  FILLING 

24.  It  is  next  necessary  to  find  the  number  of  hanks  of 
filling  in  50  yards  of  the  cloth.  As  previously  explained,  in 
order  to  find  the  amount  of  filling  in  a  cloth,  it  is  necessary 
first  to  find  the  width  at  the  reed  in  order  to  obtain  the  true 
length  of  1  pick  of  filling. 

72-sley  —  1  =  71 

71  2  =  35.5 

35.5  X  .95  =  33.725  dents  per  inch  in  the  reed 

The  number  of  dents  occupied  by  the  body  of  the  warp 
has  already  been  found  to  be  972.  To  this  number  must  be 
added  12  dents  for  the  selvage  ends,  reeded  four  per  dent, 
making  a  total  of  984  dents  occupied.  By  dividing  the  total 
number  of  dents  by  the  number  of  dents  per  inch  in  the 
reed,  the  width  at  the  reed  is  obtained:  984  -f-  33.725  =  29.18 
inches,  practically,  width  at  the  reed. 

Having  obtained  the  width  at  the  reed,  find  the  number 
of  picks  per  inch  that  the  cloth  contains.  This  can  be 
obtained  in  the  same  manner  as  were  the  number  of  ends 
per  inch  in  the  cloth,  namely,  either  by  the  use  of  a  pick 
glass  or  by  cutting  out  a  small  piece  so  many  inches  square 
and  counting  the  number  of  threads  as  they  are  pulled  from 
the  cloth.  By  counting  the  picks  in  the  sample  given  it  is 
found  to  contain  60  picks  per  inch.  Therefore,  since  1  pick 
of  filling  is  29.18  inches  long  and  there  are  60  picks  per  inch 
in  the  cloth,  in  1  inch  there  will  be  60  X  29.18  =  1,750.8  inches 
of  filling.  As  previously  explained,  there  will  be  1,750.8  yards 


12 


CLOTH  CALCULATIONS,  COTTON 


21 


of  filling  in  1  yard  of  the  cloth  and  in  50  yards  there  will  be 
50  X  1,750.8  =  87,540  yards  of  filling  in  the  cut. 

To  learn  the  number  of  hanks  divide  the  length,  in  yards, 
by  840.  87,540  h-  840  =  104.21,  number  of  hanks  of  filling 

contained  in  50  yards  of  the  cloth  similar  to  the  sample. 

The  number  of  hanks  of  warp  in  the  cut,  229.867,  added 
to  the  number  of  hanks  of  filling  in  the  cloth,  or  104.21, 
gives  334.077  as  the  total  number  of  hanks  of  yarn  in  the  cut. 


YARDS  PER  POUND 

25.  It  is  next  necessary  to  find  the  number  of  yards  per 
pound  that  this  cloth  will  contain.  For  this  purpose  it  will 
be  considered  that  a  piece  2  inches  square,  or  4  square  inches, 
is  weighed  and  found  to  weigh  3.56  grains. 

According  to  the  rule  in  Art.  19,  - 7,000  X  4 -  _  7  99 

3.56  X  27i  X  36 

or,  practically,  8  yards  per  pound,  weight  of  the  cloth. 

Since  8  yards  of  the  cloth  weighs  1  pound,  50  yards  will 
weigh  50  -f-  8  =  6.25  pounds,  which  is  the  weight  of  the  cut. 


AVERAGE  COUNTS 

26.  The  number  of  hanks  of  yarn  in  the  50  yards  of 
cloth  has  been  obtained,  also  the  weight  of  the  cut,  from 
which  it  is  possible  to  obtain  the  average  counts  of  the 
warp  yarn  and  filling  yarn  by  dividing  the  number  of  hanks 
by  the  weight. 

334.077  -T-  6.25  =  53.45s,  average  number 

27.  To  find  the  average  counts  of  yarn  in  a  piece  of  cloth: 

Rule. — Find  the  number  of  hanks  of  warp  yarn  contained  in 
the  cloth.  Add  to  this  a  certain  percentage  for  size.  Then  find 
the  number  of  hanks  of  filling  yarn  in  the  cloth  and  add  to  this 
result  the  member  of  hanks  of  warp  yarn  to  obtain  the  total  hanks 
of  yarn  in  the  cloth.  Next  find  the  weight  of  the  piece  of  cloth 
and  divide  the  total  hanks  by  this  weight.  The  result  thus 
obtained  will  be  the  average  counts. 


22 


CLOTH  CALCULATIONS,  COTTON 


§12 


COUNTS  OF  FILLING  TO  PRESERVE  YARDS 

PER  POUND 

28.  It  is  usual  in  mills  to  select  for  warp  yarn  counts 
that  are  being-  run  and  figure  for  counts  of  filling  that  will 
make  the  cloth  the  desired  weight.  This  saves  considerable 
time,  which  would  otherwise  be  consumed  in  the  changing 
over  necessary  on  the  different  machines.  A  good  method 
to  employ  when  considering  what  counts  of  warp  yarn  to  use 
to  produce  a  sample  of  cloth  is  to  have  a  number  of  samples 
of  known  counts  of  yarn  with  which  the  warp  yarn  of  the 
cloth  sample  can  be  compared;  by  this  means  an  approxi¬ 
mate  counts  can  be  obtained. 

Another  method  that  may  be  adopted,  and  one  that  may 
be  used  in  connection  with  either  the  warp  or  the  filling,  is 
to  unravel  a  certain  number  of  inches  from  the  cloth  and 
weigh  this  amount  on  grain  scales,  from  which  the  counts  of 
the  yarn  may  be  obtained  by  the  following  rule. 

To  find  the  size  of  a  given  yarn  when  the  weight  of  a 
definite  number  of  inches  is  known: 

Rule. — Multiply  the  number  of  niches  weighed  by  7,000 
( grains  in  1  pound),  and  divide  the  result  thus  obtained  by  the 
Product  of  the  weight ,  in  grains,  the  standard  number ,  and  36. 

In  case,  however,  the  counts  of  the  warp  are  figured  by 
this  method,  it  should  be  understood  that  the  weight  of  the 
yarn  found  by  weighing  the  number  of  inches  taken  from  the 
cloth,  includes  not  only  the  original  weight  of  the  yarn,  but 
also  the  weight  of  the  size  and  the  additional  weight  due 
to  contraction;  consequently,  allowances  should  be  made 
for  this. 

It  will  be  assumed  that  the  warp  yarn  of  this  cloth  has 
been  compared  with  known  counts  and  it  has  been  decided 
to  use  50s  warp.  It  is  now  necessary  to  determine  what 
weight  of  filling  must  be  used  in  order  to  have  the  cloth 
weigh  the  same,  or  8  yards  to  the  pound. 

Since  dividing  hanks  by  counts  gives  weight,  229.867  4-  50 
=  4.59  pounds,  which  is  the  weight  of  the  warp  yarn.  The 


12 


CLOTH  CALCULATIONS,  COTTON 


23 


total  weight  of  the  cut  of  cloth  is  6.25  pounds;  to  find 
the  weight  of  the  filling,  subtract  the  weight  of  the  warp 
from  the  total  weight  of  the  cut  of  cloth:  6.25  pounds  — 
4.59  pounds  =  1.66  pounds,  weight  of  filling.  Since  hanks 
divided  by  weight  gives  counts,  104.21  -4-  1.66  =  62.77,  or 
practically  63s,  counts  of  the  filling.  Therefore,  if  50s  warp 
and  63s  filling,  or  filling  sizing  slightly  heavier  than  63s,  are 
used,  the  weight  of  the  cloth  will  be,  for  all  practical  pur¬ 
poses,  the  same  as  that  of  the  sample  here  given. 

29.  To  find  the  counts  of  filling  to  preserve  weight: 

Rule. — First  find  the  weight  of  the  warp  yarn  by  dividing 
the  number  of  hanks  of  warp,  including  the  allowa?ice  for  size , 
by  the  counts  of  warp.  Subtract  the  result  tluis  obtained  from 
the  total  weight  of  the  cut  of  cloth  and  divide  this  result  into 
the  hanks  of  filling  in  the  cut.  The  result  will  be  the  necessary 
counts  of  filling. 

If  it  is  desired  to  find  the  counts  of  warp  to  preserve  the 
yards  per  pound,  this  rule  may  be  used  by  simply  substitu¬ 
ting  the  word  warp  for  filling,  and  vice  versa. 


AVERAGE  SEEY 

30.  One  other  specification,  which  has  not  been  referred 
to  but  which  must  frequently  be  obtained,  especially  in  cloths 
such  as  the  one  under  description,  is  the  average  sley,  or 
in  other  words,  the  average  number  of  ends  per  inch  that 
the  cloth  contains.  The  most  accurate  method  of  obtaining 
this  particular  is  to  divide  the  total  number  of  ends  in 
the  warp,  exclusive  of  selvages,  by  its  width.  3,264  -4-  27 
=  121,  average  sley,  nearly.  Thus  the  average  sley  of  the 
cloth  being  dealt  with  is  121,  while  a  72-sley  reed  is  to  be 
used;  i.  e.,  a  reed  that  if  drawn  two  per  dent  will  produce  a 
72-sley  cloth.  These  particulars  are  often  written  72/121. 

All  the  necessary  particulars  for  the  reproduction  of  this 
sample  have  been  obtained  and,  according  to  the  method 
previously  stated,  they  are  shown  as  follows:  72/121  X  60 
—  27|  in.  —  50s  warp  —  63s  filling  —  8-yard  cloth. 


24 


CLOTH  CALCULATIONS,  COTTON 


§12 


SHORT  RULES  USED  IN  OBTAINING  PARTICULARS 

OF  CLOTH  SAMPLES 

31.  In  the  calculations  that  have  so  far  been  given 
regarding  this  sample  of  cloth,  prominence  has  been  given 
to  the  reason  for  procedure  rather  than  to  the  shortest  pos¬ 
sible  methods,  since  it  is  impossible  to  attach  too  much 
importance  to  the  advantage  derived  from  working  by  rea¬ 
son  rather  than  by  rule.  However,  in  many  cases  shorter 
methods  than  those  given  will  be  found  to  be  almost  as 
accurate,  and  yet  in  many  instances,  especially  where  the 
contraction  of  the  different  yarns  in  the  warp  is  found  to 
differ,  the  particulars  of  a  piece  of  cloth  must  be  worked 
out,  each  item  by  itself,  in  order  to  obtain  results  at  all 
accurate. 

32.  To  find  the  average  number  of  yarn  in  a  cloth  of 
ordinary  construction: 

Rule. — Add  the  sley  and  the  pick  together;  multiply  this 
result  by  the  width  and  the  result  Unis  obtained  by  the  yards 
per  pound  and  divide  this  result  by  750.  The  answer  will  be 
the  average  number  of  the  yarns. 


The  standard  840  has  to  be  reduced  by  a  certain  number 
of  yards,  in  order  to  make  up  for  the  changes  in  weight 
and  length  caused  by  sizing  and  contraction.  In  general, 
90  yards  is  deducted,  leaving  750  as  the  standard  number  of 
yards  to  the  pound. 


Example. — It  is  desired  to  find  the  average  number  of  a  cloth  con¬ 
taining  60  ends  and  66  picks  per  inch,  the  cloth  being  30  inches  wide 
and  weighing  5  yards  per  pound. 


Solution.—  60  +  66  =  126;  126  X  30  =  3,780 
3,780  X  5  =  18,900;  18,900  -4-  750  =  25.2s,  average  number, 
or  60  +  66  =  126 


126  X  30  X  5 
750 


25.2s,  average  number.  Ans. 


Ans. 


If  this  supposed  piece  of  cloth  is  worked  out  in  a  similar 
manner  to  that  adopted  when  dealing  with  the  sample  of 


§12 


CLOTH  CALCULATIONS,  COTTON 


25 


cloth  in  this  Section  it  will  be  seen  that  the  answers  are  prac¬ 
tically  the  same. 

33.  Another  rule  that  will  be  found  accurate  when  deal¬ 
ing  with  cloths  of  ordinary  construction  is  that  of  obtaining 
the  filling  required  to  preserve  the  weight  of  the  cloth  when 
the  average  number  of  the  yarns  in  the  cloth  and  the  counts 
of  the  warp  are  known. 

Rule. — Add  the  s ley  and  the  pick  together  and  divide  by  the 
average  number.  Divide  the  sley  by  the  counts  of  the  warp. 
Subtract  the  result  obtained  in  the  second  instance  from  the  result 
obtained  in  the  first  and  divide  the  result  tlnis  obtained  into  the 
picks  per  inch.  The  answer  will  be  the  counts  of  the  filling 
required. 

Example.— With  the  particulars  the  same  as  in  Art.  32,  but  taking 
22s  as  the  counts  of  the  warp,  find  the  counts  of  filling  required  to  be 
used  to  preserve  the  weight  of  the  cloth. 

Solution.—  60  +  66  =  126;  126  -t-  25.2  =  5 
60  -f-  22  =  2.72;  5  -  2.72  =  2.28 

66  -s-  2.28  =  28.94s,  counts  of  filling  required  to  preserve  weight.  Ans. 

34.  When  the  warp  and  the  filling,  or  the  filling  alone, 
contain  different  counts  of  yarn,  the  average  counts  of  the 
filling  must  first  of  all  be  found  in  the  same  manner  as  was 
employed  in  previous  examples  when  finding  the  counts 
of  filling  required  to  produce  cloth  of  a  given  weight.  Then, 
with  the  counts  of  one  of  the  kinds  of  filling  known,  find  the 
counts  of  the  other  filling  required  to  produce  cloth  of  the 
given  weight. 

Rule. — Divide  the  total  number  of  picks  in  the  pattern  by  the 
average  counts  of  the  fillmg.  Also  divide  the  number  of  picks  of 
the  known  counts  of  filling  by  its  counts.  Subtract  the  result 
obtained  in  the  second  instance  from  the  result  obtained  in  the 
first ,  and  divide  the  difference  into  the  number  of  picks  of  the 
unknown  counts. 

Example. — A  piece  of  cloth,  64  X  64,  27  inches  wide,  has  the  warp 
and  filling  arranged  46  ends  of  fine  and  3  ends  of  cord.  The  counts 
of  the  fine  yarn  in  the  warp  are  30s  and  of  the  cord  10s.  If  the  cloth 


26  CLOTH  CALCULATIONS,  COTTON  §12 


weighs  6.4  yards  to  the  pound,  what  counts  of  fine  filling  must  be 
employed  to  preserve  the  yards  per  pound? 

Solution. — First  find  the  average  counts  of  the  warp. 


46  4-  30  =  1.53 
3  10  =  .30 

1.83 

49  -r-  1.83  =  27s,  nearly,  average  counts  of  warp 
Next  find  the  average  counts  of  warp  and  filling. 


128  X  27  X  6.4 
750 


64  +  64  =  128 

29.5s,  nearly,  average  counts  of  warp  and  filling 


Next  find  the  average  counts  of  filling. 


64  +  64  =  128;  128  --  29.5  =  4.34 
64  +27  =  2.37;  4.34  -  2.37  =  1.97 
64  -f-  1.97  =  32.5s,  average  counts  of  filling 

The  question  now  is  to  find  the  counts  of  the  cord  and  the  fine  yarn 
in  the  filling  to  preserve  the  yards  per  pound,  the  average  counts  of 
the  filling  and  the  arrangement  of  the  yarn  in  the  filling  being  known. 
In  cases  of  this  kind  it  would  be  unlikely  that  a  mill  would  employ 
different  counts  of  cord  in  both  warp  and  filling,  consequently  it  would 
be  safe  to  assume  the  counts  of  the  cord  in  the  filling  to  be  the  same 
as  that  in  the  warp,  after  which  it  would  only  be  necessary  to  find  the 
counts  of  the  fine  filling. 

49  -r  32.5  =  1.5 
3  -r  10  =  .3 

1.2 

46  -r  1.2  =  38s,  counts  of  fine  filling.  Ans. 


CLOTH  CALCULATIONS, 
WOOLEN  AND  WORSTED 


CALCULATIONS  NECESSARY  FOR 
CLOTH  PRODUCTION 


INTRODUCTION 

1.  Importance  of  Cloth  Calculations. — The  calcula¬ 
tions  in  mill  work  that  deal  with  the  construction  of  fabrics, 
known  as  cloth  calculations,  are  by  no  means  the  least 
important  of  the  many  calculations  dealing  with  woolen  and 
worsted  mill  processes.  Important  alike  to  the  treasurer 
and  superintendent,  and  of  indispensable  value  to  the 
designer,  they  play  a  leading  part  in  governing  and  standard¬ 
izing  the  construction  of  all  woven  fabrics.  When  laying 
out  a  fabric,  or  when  analyzing  a  sample  of  cloth  with  a 
view  to  its  reproduction,  not  only  must  the  many  details 
of  its  structure  be  ascertained,  but  it  must  also  be  learned 
whether  it  is  possible  for  the  mill  to  make  the  cloth  with  the 
machinery  with  which  it  is  equipped.  The  term  cloth  calcu¬ 
lations  is  sufficiently  wide  in  its  scope  to  include  a  number 
of  topics,  each  of  which  must  be  thoroughly  understood 
before  a  comprehensive  knowledge  of  the  subject  can  be 
obtained.  In  order  that  the  student  may  obtain  such  a 
knowledge,  the  different  calculations  usually  met  with  in 
connection  with  a  woolen  or  worsted  fabric  will  be  explained 


For  notice  of  copyright,  see  page  immediately  following  the  title  page 

l  13 


2 


CLOTH  CALCULATIONS, 


13 


first,  giving  rules  when  necessary,  and  finally  applied  to  a 
given  sample  of  cloth. 

Since  the  methods  of  cloth  calculation  that  prevail  in 
England  and  on  the  continent  of  Europe  differ  so  widely 
from  those  used  in  the  United  States,  it  has  been  deemed 
best  to  confine  the  explanations  and  rules  to  American 
methods  in  order  to  avoid  the  confusion  of  too  many 
systems  and  their  consequent  explanation. 


CONSTRUCTION  OF  FABRICS 

2.  Woven  fabrics  are  constructed  by  the  interlacing  of 
two  or  more  systems  of  yarns  technically  known  as  the  warp 
and  filling.  The  warp  is  the  system  of  yarn  that  lies  in  the 
direction  of  the  length  of  the  goods,  while  the  filling  is  the 
yarn  that  crosses  the  warp  at  right  angles  and  lies  in  the  direc¬ 
tion  of  the  width  of  the  cloth.  A  single  thread  of  the  warp 
yarn  is  known  as  an  end,  and  a  single  thread  of  the  filling 
yarn  as  a  pick.  The  texture  of  a  woolen  or  worsted  cloth 
is  designated  as  so  many  ends  and  so  many  picks  per  inch. 
Therefore,  it  might  be  said  that  a  certain  cloth  had  72  ends 
and  60  picks  per  inch.  Sometimes  this  is  written  simply 
72  X  60. 

After  the  warp  for  a  fabric  is  made,  it  is  wound  on  a 
loom  beam  and  the  separate  ends  drawn  through  what  are 
known  as  harnesses,  each  end,  however,  being  drawn 
through  only  one  harness  in  the  production  of  nearly  all 
fabrics.  After  the  warp  is  drawn  through  the  harnesses,  it 
is  also  drawn  through  the^reed.  The  warp  is  then  ready  to 
be  placed  in  the  loom.  The  harnesses  are  attached  to  the 
mechanism  of  the  loom,  by  means  of  which  they  are  raised 
and  lowered  independently  of  each  other,  and,  since  some 
harnesses  are  up  while  others  are  down,  a  division  of  the 
warp  is  effected.  This  division  is  known  as  a  shed,  and  it 
is  through  the  shed  that  the  shuttle  carrying  the  filling  is 
passed,  thus  interlacing  the  filling  with  the  warp.  Each  pick 
of  filling  is  beaten  into  the  cloth  by  the  reed,  which  is  fastened 
to  an  oscillating  portion  of  the  loom  known  as  the  lay. 


13 


WOOLEN  AND  WORSTED 


3 


WEIGHT  OF  CLOTH 

3.  The  weight  of  woolen  or  worsted  cloth  is  designated 
by  the  ounces  per  yard;  thus,  a  cloth  that  weighs  12  ounces 
per  yard  is  called  a  12-ounce  cloth.  This  weight  is  not  the 
weight  of  1  square  yard  of  cloth  (unless  the  cloth  happens 
to  be  36  inches  wide),  but  of  1  linear  yard,  measured  in  the 
direction  of  the  warp  of  the  goods.  The  width  depends  on 
the  width  of  the  goods,  which  may  be  28,  72,  or  any  other 
number  of  inches,  as  the  case  may  be.  Goods  that  are 
28  inches  wide  are  known  as  single-width  goods,  56-inch 
goods  are  known  as  double-width.  On  other  widths  the 
width  of  the  goods  is  designated  in  inches. 

The  width  of  woolen  and  worsted  fabrics  is  often  desig¬ 
nated  by  such  expressions  as  i  goods,  f  goods,  etc.  While 
it  is  sometimes  stated  that  these  expressions  mean  that 
the  cloth  is  f  yard  or  f  yards  in  width,  as  the  case  may  be, 
this  is  not  strictly  correct.  The  expressions  are  based  on 
the  Scotch  ell,  which  is  equal  to  37.09  inches;  and  thus 
the  expression  f  goods  really  means  that  the  cloth  is  27.81 
inches  wide,  and  f  goods  that  the  cloth  is  55.62  inches 
wide.  Commercially,  however,  the  generally  accepted  width 
for  i  goods  in  all  markets  is  28  inches  and  for  i  goods 
56  inches. 

In  the  cotton  trade  the  weight  of  cloth  is  not  usually 
designated  by  the  ounces  per  linear  yard,  but  by  the  linear 
yards  per  pound.  Thus,  in  a  cotton  mill,  a  2-ounce  cloth 
would  be  known  as  an  8-yard  cloth,  since  if  1  yard  weighs 
2  ounces,  there  are  8  yards  in  1  pound  (16  ounces).  This 
applies  more  particularly  to  print  cloths  and  ginghams;  in 
heavier  cotton  goods,  it  is  customary  to  designate  the  weight 
of  the  cloth  in  ounces  per  yard,  as  with  woolen  and  worsted 
fabrics.  Cotton  duck  is  designated  by  the  ounces  per 
square  yard,  and  is  made  in  various  weights  and  widths. 


4  CLOTH  CALCULATIONS,  §13 


HARNESS  CALCULATIONS 


*8  ,  ,  r  »  ,  ..  -  -  » 

K? 

*  S 


4.  The  harnesses,  through  which  the  ends  of  the  warp 
are  drawn,  are  composed  of  a  framework  of  wood  through 
which  two  flat  steel  rods  are  passed,  one  near  the  top  and 
one  near  the  bottom.  These  rods  are  known  by  different 
names,  such  as  heddle  bars,  heddle  rods,  etc.  On  the  heddle 
bars  are  placed  wire  heddles,  which  have  eyes  in  the  centers 
through  which  the  warp  ends  are  drawn.  Whenever  a  new 
warp  is  drawn  in,  it  becomes  necessary  to  find  the  number 
of  heddles  that  must  be  placed  on  each  harness,  in  order  to 

make  the  harnesses 
correspond  to  the 
drawing-in  draft  and 
warp.  To  perform 
such  a  calculation  it 
is  necessary  to  know 
the  manner  in  which 
the  ends  are  to  be 
drawn  through  the 
harnesses,  or,  in 
other  words,  the  drawing-in  draft.  A  drawing-in,  or 
harness,  draft  will  be  readily  understood  by  referring  to 
the  draft  in  Fig.  1,  which  shows  through  which  harness  each 
end  of  the  warp  is  drawn.  It  will  be  noticed  that  the  first 
end  is  drawn  through  the  first  harness;  the  second  end 
through  the  second  harness;  the  third  end  through  the  first 
harness;  and  so  on  throughout  the  10  ends  that  constitute 
one  repeat  of  the  drawing-in  draft.  As  this  is  one  repeat  of 
the  draft,  all  of  the  other  ends  of  the  warp  are  drawn  through 
the  harnesses  in  a  similar  manner. 


4th  Harness 
3rd  , , 
2nd  ,, 

1  st  , « 


4 

3 

3 

2 

2 

2 

2 

1 

1 

1 

Fig.  1 


5.  To  find  the  number  of  heddles  required  on  each  har¬ 
ness,  the  drawing-in  draft  and  the  number  of  ends  in  the 
warp  being  known: 

Rule. — Find  the  number  of  repeats  of  the  drawing-in  draft  in 
the  warp  by  dividing  the  total  number  of  ends  in  the  warp  by  the 
number  of  ends  in  one  repeat  of  the  draft.  Multiply  the  quotient 


§13 


WOOLEN  AND  WORSTED 


5 


thus  obtained  by  the  number  oh  ends  drawn  in  on  each  harness  in 
one  repeat  of  the  drawing-in  draft.  The  results  obtained  will ,  in 
each  instance ,  be  the  number  of  lieddles  required  for  that  particular 
harness. 

Example. — A  warp  containing  2,400  ends  is  drawn  in  on  four  har¬ 
nesses  according  to  the  draft  given  in  Fig.  1;  how  many  heddles  are 
required  on  each  harness? 

Solution. — Counting  across  the  draft,  it  will  be  seen  that  there  are 
10  ends  in  one  repeat  of  the  drawing-in  draft,  and  that  3  ends  are 
drawn  in  on  the  first  harness,  4  ends  on  the  second  harness,  2  ends  on 
the  third  harness,  and  1  end  on  the  fourth  harness. 

2,400  -5-  10  =  240  repeats  of  the  draft 
240  X  3  =  720  heddles  on  the  first  harness 
240  X  4  =  960  heddles  on  the  second  harness 
240  X  2  =  480  heddles  on  the  third  harness  nS’ 

240  X  1  =  240  heddles  on  the  fourth  harness  . 

As  a  rule,  a  few  extra  heddles  are  placed  on  each  harness,  in  order 
to  meet  any  additional  requirements  that  may  occur,  such  as  broken 
heddles,  selvages,  etc. 

If  2  or  more  ends  of  the  warp  are  drawn  through  each 
heddle,  as  is  sometimes  done  when  weaving  certain  fabrics, 
the  number  of  heddles  required  will  be  correspondingly 
reduced  and  the  results  obtained  by  the  above  rule  must  be 
divided  by  the  number  of  ends  drawn  in  a  heddle.  This 
may  occur  on  all  the  harnesses  in  some  cases,  and  in  others 
on  only  one  or  two  harnesses. 


REED  CALCULATIONS 

6.  Before  the  cloth  is  ready  to  be  woven,  the  warp  must 
be  drawn  not  only  through  the  harnesses,  but  also  through 
the  reed.  Since  the  reed  plays  a  very  important  part,  not 
only  in  the  weaving,  but  also  in  the  calculations  connected 
with  cloth,  it  becomes  necessary  to  give  a  short  description 
of  reeds  and  the  terms  used  in  connection  with  them.  Reeds 
are  constructed  of  thin,  flat  strips  of  steel  set  edgewise  in  two 
pieces  of  wood  called  ribs,  the  ribs  being  wound  with  waxed 
cord  in  order  to  make  the  whole  reed  firm. 


6  CLOTH  CALCULATIONS,  §13 

7.  Number  of  tlie  Reed. — The  space  between  two 
adjoining  strips  of  steel  in  a  reed  is  called  a  split,  or  dent, 
the  latter  term  being  the  one  generally  employed  in  Ameri¬ 
can  mills.  The  number  of  dents  in  a  given  space  determines 
the  number  of  the  reed.  The  general  custom  is  to  use 
1  inch  for  this  given  space;  thus,  a  15s  reed  contains  15  dents 
per  inch.  In  the  English  woolen  trade,  I  yard,  or  9  inches, 
is  sometimes  used;  thus,  a  135s  reed  by  this  system  is  the 
same  as  a  15s  reed  by  the  American  system.  Sometimes, 
especially  in  the  cotton  trade,  the  number  of  dents  in  a  given 
number  of  inches  is  used  to  designate  the  number  of  the  reed; 
thus,  a  1,200-30-reed,  which  indicates  that  there  are  1,200 
dents  in  30  inches,  is  equal  to  a  40s. 

Reeds  as  sent  out  by  the  manufacturers  are  always 
marked  in  one  of  the  two  methods  indicated  above;  that  is, 
either  according  to  the  number  of  dents  per  inch  or  the 
number  of  dents  in  a  certain  number  of  inches.  However, 
reeds  are  sold  by  the  bier.  The  bier,  as  applied  to  reeds, 
means  20  dents;  consequently,  when  the  price  of  a  reed  is 
quoted  at  so  much  per  bier,  it  means  so  much  for  every 
20  dents. 

The  coarser  a  reed,  to  a  certain  extent,  the  less  friction 
there  will  be  on  the  warp;  and  on  the  other  hand,  the  finer 
the  reed,  the  smoother  will  be  the  fabric.  Sometimes  if  too 
many  warp  ends  are  drawn  per  dent  of  the  reed,  they  will 
roll  or  ride  each  other.  This  may  usually  be  remedied  by 
using  a  finer  reed  and  drawing  fewer  ends  per  dent. 

The  number  of  inches  that  a  warp  occupies  in  the  reed  is 
called  the  set  of  the  cloth;  thus,  a  piece  of  goods  would  be 
said  to  be  set  66  inches  wide  in  the  reed;  this  is  also  called 
the  width  of  the  cloth  in  the  loom.  Woolen  and  worsted, 
and  in  fact  most  fabrics,  except  in  special  cases,  have  to 
be  reeded  wider  than  the  width  of  the  finished  goods,  owing 
to  the  contraction  of  the  goods  after  they  are  taken  from  the 
loom  and  the  shrinkage  of  the  cloth  in  finishing. 

The  height  of  the  reed,  or  the  length  of  the  reed  wires  that 
separate  the  dents  measured  between  the  ribs,  is  governed 
by  the  class  of  fabric  to  be  woven  and  the  kind  of  loom  for 


§13 


WOOLEN  AND  WORSTED 


7 


which  it  is  designed.  For  instance,  a  3i-inch  reed  is  high 
enough  for  most  classes  of  cotton  goods,  while  woolens  and 
worsteds  require  from  a  4i-  to  a  6-inch  reed.  Carpets  and 
other  heavy  fabrics  of  a  similar  nature  require  a  higher  reed 
— often  as  high  as  8  or  9  inches. 

There  are  a  few  calculations  in  regard  to  reeds  that 
should  be  thoroughly  understood. 

8.  To  find  the  width  of  the  cloth  in  the  reed  when  the 
number  of  ends  per  dent,  the  number  of  the  reed,  and  the 
total  number  of  ends  in  the  warp  are  known: 

Rule. — Divide  the  total  number  of  ends  in  the  warp  by  the 
number  of  the  reed  multiplied  by  the  ends  per  dent. 

Example. — A  worsted  warp  is  drawn  in  three  per  dent  in  a  20s  reed; 
the  warp  contains  1,920  ends;  how  wide  is  it  in  the  reed? 

_  1,920 

Solution. —  o?w~q  =  ^2  in.  Ans. 

JU  X  o 

9.  Selvages. — One  point  that  should  be  noted  here  is 
that  on  each  side  of  practically  all  cloths  there  are  additional 
ends  drawn  through  the  harnesses,  or  through  a  special  motion 
that  has  the  same  function  as  harnesses,  and  also  through  the 
reed  at  each  side  of  the  true  warp.  These  are  known  as 
selvage ,  or  listing ,  ends ,  and,  when  interwoven  with  the  filling, 
form  a  strong  tape,  or  strip,  on  each  edge  of  the  cloth,  which 
is  known  as  a  selvage,  or  listing.  A  larger  number  of 
selvage  ends  are  usually  drawn  in  each  dent  than  in  the 
body  of  the  warp,  in  many  cases  twice  as  many.  This  is  for 
the  purpose  of  producing  firm,  strong  edges  on  the  cloth. 
Selvages  vary  from  i  to  li  inches  in  width,  depending  on 
the  class  of  fabric  that  is  being  woven.  Listing  ends  must 
always  be  taken  into  consideration  when  finding  the  width 
of  a  fabric  in  the  reed  unless  the  width  inside  selvages 
is  designated. 

Suppose  that  in  the  example  in  Art.  8  there  are  30  ends  of 
listing  drawn  six  in  a  dent  on  each  side  of  the  cloth.  Then, 
to  find  the  total  width  in  the  reed,  the  width  occupied  by  the 
selvages  must  be  added  to  the  width  of  the  warp  in  the  reed. 
In  this  case  there  are  30  ends,  on  each  side,  or  60  selvage 


5 


CLOTH  CALCULATIONS, 


§13 


ends,  which,  drawn  six  in  a  dent  in  a  20s  reed,  equals  .5  inch 

(  =  ,5).  This,  added  to  the  width  of  the  warp  in  the 

\20  X  6  / 

reed,  or  32  inches,  equals  32.5  inches,  total  width  in  reed 
including  selvages. 

10.  To  find  the  total  number  of  ends  in  a  warp  when  the 
width  in  the  reed  and  the  ends  per  dent  are  known: 

Rule. — Multiply  the  number  of  the  reed  by  the  ends  per  dent 
and  by  the  width  in  the  reed  and  to  this  result  add  a  certain 
member  of  e?ids  for  selvages. 

Example. — A  warp  is  reeded  four  per  dent  in  a  12s  reed  and  set 
34  inches  wide  inside  of  selvages;  how  many  ends  are  there  in  the  warp 
if  20  ends  are  used  for  listing  on  each  side  of  the  cloth,  or  40  ends  in  all? 

Solution. —  12  X  4  X  34  =  1,632;  1,632  +  40  =  1,672  ends.  Ans. 


11.  To  find  the  reed  when  the  width  in  the  loom,  the 
number  of  ends  in  the  warp,  and  the  ends  per  dent 
are  known: 


Rule. — Divide  the  member  of  e?ids  in  the  warp  by  the  width 
in  the  loom  multiplied  by  the  ends  per  dent. 


Example. — A  warp  contains  2,400  ends  and  is  to  be  set  30  inches 
wide  in  the  loom;  what  reed  is  required  to  reed  it  four  per  dent? 


Solution. — 


2,400  on  a 
3tT><4  =  20s  reed' 


Ans. 


12.  When  cloth  is  unevenly  reeded  it  is  more  difficult  to 
find  the  required  reed  than  with  evenly  reeded  cloth.  In 
cloth  of  this  description,  the  reeding  of  the  warp  is  usually 
arranged  in  a  definite  order  in  conjunction  with  the  pattern 
of  the  warp,  which  is  generally,  although  not  always,  com¬ 
posed  of  yarns  of  different  counts. 

13.  To  find  the  reed  to  use  for  an  unevenly  reeded  cloth: 

Rule. — Multiply  the  dents  in  one  pattern  by  the  patterns  in 
the  warp  and  divide  by  the  width  in  the  loom. 


WOOLEN  AND  WORSTED 


9 


§13 

Example. — A  warp  contains  3,920  ends  and  is  to  be  reeded  62  inches 
wide  in  the  loom;  what  number  of  reed  is  required  if  the  warp  is 
dressed  and  reeded  as  follows? 


4 

white 

drawn  in 

1  dent 

1 

white 

1 

1 

tan 

white 

>  drawn  in 

1  dent 

1 

tan 

4 

white 

drawn  in 

1  dent 

12 

tan 

drawn  in 

6  dents 

4 

white 

drawn  in 

1  dent 

1 

white1 

1 

1 

tan 

white 

*  drawn  in 

2  dents 

1 

tan 

12 

white 

drawn  in 

6  dents 

4 

blue 

drawn  in 

1  dent 

1 

tan 

2 

white 

drawn  in 

2  dents 

1 

tan 

144 

tan 

drawn  in 

7  2  dents 

19  6  ends  in  pattern  9  3  dents  in  pattern 

Note.— The  above  is  known  as  a  reed  draft. 

Solution. —  3,920  -p  196  =  20  patterns  in  warp 

=  30s  reed.  Ans. 


CONTRACTION  IN  WEAVING 

14.  Fabrics  contract  in  two  ways  during  the  weaving 
process.  They  contract  in  length;  that  is,  to  weave  one  cut 
of  50  yards  of  cloth,  the  warp  must  be  longer  than  50  yards 
because  of  the  contraction  in  interlacing  with  the  filling.  As 
already  explained,  all  cloths  are  formed  by  the  interlacing  of 
two  series  of  yarns — the  warp  and  the  filling — and,  since  in 
interlacing  in  weaving,  the  two  series  of  yarns  bend  around 
each  other  to  a  greater  or  less  extent,  it  naturally  follows 
that  a  piece  of  cloth  will  not  be  so  long  as  the  warp  from 
which  it  was  woven.  There  will  be  found  many  causes  that 
affect  the  amount  of  contraction  of  the  warp  in  weaving. 
The  tension  of  the  warp  during  weaving  will  affect  it  to  a 


10 


CLOTH  CALCULATIONS, 


§13 


slight  extent.  The  comparative  counts  of  the  warp  and 
filling  also  have  an  influence,  since  it  will  readily  be  seen 
that,  if  the  warp  is  made  of  coarser  yarn  than  the  filling,  it 
will  not  be  deflected  to  so  great  an  extent  and  therefore 
will  not  contract  so  much.  The  contraction  of  the  warp  in 
weaving  is  known  as  take-iip  in  weaving. 

The  cloth  also  contracts  in  the  width,  or  in  the  direction  of 
the  filling;  that  is,  the  woven  cloth  is  narrower  than  the  cloth 
is  reeded.  This  latter  depends  largely  on  the  character  of 
the  filling,  whether  coarse  or  fine,  or  hard  or  soft  twist. 

The  contraction  of  goods  in  weaving,  in  general,  is  largely 
influenced  by  the  class  of  weave  that  is  used,  or  in  other 
words  the  manner  of  interlacing  the  warp  and  filling;  thus,  it 
will  be  seen  that  a  weave  that  interlaces  only  once  in  5  ends 
or  5  picks,  like  a  satin  weave,  will  not  contract  the  same  as  a 
weave  that  interlaces  at  every  end  and  pick,  like  a  plain 
weave.  Often  certain  weaves  will  greatly  increase  the  con¬ 
traction  of  one  system  of  yarn,  either  warp  or  filling,  and 
correspondingly  decrease  the  contraction  of  the  other.  The 
contraction  of-  cloth  in  weaving  varies  from  4  to  15  per  cent. 
No  definite  rule  can  be  given  for  this  contraction  with  woolen 
and  worsted  goods,  and  the  value  of  minute  records  of  each 
piece  of  cloth  that  the  mill  makes  cannot  be  overestimated. 

The  contraction  in  weaving  must  not  be  confounded  by 
the  student  with  the  shrinkage  of  the  cloth  in  finishing.  This 
latter  varies  with  the  class  of  goods  from  10  to  15  per  cent, 
for  cassimeres  up  to  from  25  to  30  per  cent,  for  beavers 
and  kerseys. 

15.  To  find  the  length  of  warp  required  to  weave  a 
required  length  of  cloth,  the  take-up  of  the  warp  in  weaving 
being  known: 

Rule. — Divide  the  required  length  of  cloth  by  100  per  cent, 
minus  the  percentage  of  take-up  in  weaving. 

Example. — How  many  yards  of  warn  are  necessary  to  weave 
72  yards  of  cloth  if  the  take-up  in  weaving  is  4  per  cent.? 

Solution. —  100  per  cent.  —  4  per  cent.  =  96  per  cent. 

72  -p  .96  =  75  yd.  Ans. 


13 


WOOLEN  AND  WORSTED 


11 


Explanation. — It  would  seem  in  the  above  example  that 
all  that  is  necessary  is  to  find  4  per  cent,  of  72  yards  and 
add  it,  but  it  must  be  remembered  that  it  is  the  warp 
that  takes  up  4  per  cent.,  not  the  cloth.  The  length  of  the 
cloth  (72  yards)  is  the  length  of  the  warp  minus  4  per  cent.; 
therefore,  it  is  96  per  cent,  of  the  original  length  of  warp. 
That  this  example  is  correct  may  be  seen  by  finding  4  per 
cent,  of  the  length  of  the  warp  (75  yards)  and  subtracting 
the  length  obtained  (3  yards)  from  it. 

16.  In  making  allowances  for  various  percentages  when 
figuring  any  contraction  or  shrinkage  of  cloth  or  yarn  or 
other  results  dependent  on  such  contraction  or  shrinkage,  in 
case  of  any  doubt  as  to  the  proper  method  of  applying  the 
percentage  the  following  rules  should  be  borne  in  mind: 

Rule  I. —  When  figuring  forwards  in  the  direction  of  the  process 
of  manufacture,  as  for  instance ,  from  yarn  to  cloth ,  or  from  woven 
cloth  to  finished  cloth ,  if  the  result  is  to  be  increased ,  multiply  by 
100  per  cent,  plus  the  percentage  to  be  taken;  or  if  the  result  is  to 
be  decreased,  by  100  per  cent,  minus  the  percentage  to  be  taken. 

Rule  II. —  When  figuring  in  the  opposite  direction  to  the  proc¬ 
ess  of  manufacture,  as  for  instance,  from  finished  cloth  to  woven 
cloth,  or  from  woven  cloth  to  yarn,  divide,  if  the  resjilt  is  to  be 
increased,  by  100  per  cent,  mums  the  percentage  to  be  taken;  or 
if  the  result  is  to  be  decreased,  by  100  per  cent,  plus  the  per¬ 
centage  taken.  _ 

SHRINKAGE  IN  FINISHING 

17.  The  shrinkage  of  a  cloth  in  finishing  is  entirely  dis¬ 
tinct  from  the  contraction  of  the  cloth  in  weaving,  and  is  an 
item  that  can  only  be  determined  by  the  experience  and 
judgment  that  is  obtained  by  a  familiarity  with  different 
fabrics,  the  methods  employed  in  finishing  them,  and  their 
peculiarities  in  finishing.  After  close  observation  it  will  be 
possible  to  tell  about  how  much  each  fabric  made  in  the 
mill  will  shrink  in  the  finishing  process.  The  shrinkage,  of 
course,  varies  with  the  stock  entering  into  the  goods,  the 
kind  of  cloth  desired,  and  the  finishing  process.  Goods 


12 


CLOTH  CALCULATIONS, 


§13 


should  be  reeded  wide  enough  to  allow  for  the  shrinkage 
during  finishing.  It  should  be  understood,  however,  that 
although  due  consideration  of  the  probable  shrinkage  is  nec¬ 
essary,  there  is  some  leeway,  and  the  finisher  will  finish  the 
goods  to  the  width  and  weight  required,  within  reasonable 
limits. 

Woolen  goods  shrink  more  than  worsted,  and  consequently 
should  be  reeded  wider  and  warped  longer  for  the  same  width 
and  length  of  cloth.  Goods  that  are  fulled  shrink  more  than 
goods  that  are  not  fulled.  Heavy  woolen  goods  usually 
are  heavily  fulled,  often  being  triple  milled,  and  will  shrink 
as  high  as  30  per  cent.,  averaging  from  25  to  30  per  cent. 
For  light-weight,  fulled  woolens  an  allowance  for  shrinkage 
in  width  of  from  12^  to  18  per  cent,  is  necessary,  while  if 
not  fulled,  a  smaller  allowance,  say  from  10  to  15  per  cent., 
is  sufficient.  For  light-weight  worsted  goods  with  a  clear 
finish,  from  8  to  12i  per  cent,  is  a  sufficient  allowance  for 
shrinkage  from  reed  to  finished  cloth,  while  if  fulled  (which 
is  rarely  done)  from  12i  to  15  per  cent,  should  be  allowed. 
For  heavy-weight  worsted  goods  with  a  clear  finish,  from 
12i  to  15  per  cent,  should  be  allowed  for  the  shrinkage  in 
width,  while  if  the  goods  are  fulled,  an  allowance  of  from 
15  to  20  per  cent.,  or  even  more  in  some  cases,  should 
be  made. 

As  a  general  rule,  fabrics  do  not  shrink  as  much  in  length  as 
as  in  width,  especially  those  that  are  not  fulled,  the  action  in 
passing  through  the  finishing  machinery  being  to  keep  the 
goods  stretched  in  length.  This,  however,  is  not  a  universal 
rule.  When  a  fabric  shrinks  in  length,  the  counts,  or  size,  of 
ihe  warp  yarn  is  made  coarser  and  the  weight  of  the  cloth  per 
yard  is  increased.  A  shrinkage  in  width  does  not  result  in 
an  increase  in  weight,  but  does  make  the  counts  of  the  filling 
coarser,  since  the  width  of  the  cloth  is  made  narrower  and 
the  length  of  the  filling  consequently  shorter,  while  the 
weight  remains  the  same.  The  gain  in  weight  that  a  fabric 
obtains  in  shrinking  in  length,  and  also  the  increase  in  the 
counts  of  the  warp  and  filling,  is  somewhat  altered  by  the 
loss  of  weight  in  finishing,  which  occurs  from  scouring  out 


13 


WOOLEN  AND  WORSTED 


13 


the  oil  and  grease  in  the  cloth,  as  well  as  the  loss  in  certain 
processes,  such  as  napping,  shearing,  etc.  In  the  case  of 
worsted  goods,  this  loss  will  about  counterbalance  the 
weight  gained  by  the  shrinkage  of  the  cloth  in  length,  so 
that  the  counts  of  warp  and  filling  and  the  weight  per  yard 
of  the  finished  goods  are  practically  the  same  as  from  the 
loom.  Woolen  goods  if  shrunk  to  a  considerable  extent  in 
length  will  weigh  more  per  yard  after  finishing  than  before, 
but  if  not  shrunk  greatly  in  length,  will  be  lighter  when 
finished,  owing  to  the  loss  in  weight  in  finishing,  due  to  the 
cleansing  of  the  cloth  from  grease,  etc. 


18.  To  find  the  original  size  of  a  yarn  when  the  counts 
of  the  yarn  in  the  finished  cloth,  the  total  shrinkage,  and  the 
loss  in  finishing  are  known: 


Rule. — Divide  the  finished  cozints  of  the  yarn ,  multiplied 
by  100,  by  100  minus  the  difference  between  the  total  shrinkage 
and  the  loss  in  finishing. 

Example  1. — If  the  warp  in  a  piece  of  cloth  has  shrunk  20  percent, 
in  length  from  the  beam  to  the  finished  cloth,  and  the  goods  have 
lost  10  per  cent,  in  finishing,  what  were  the  original  counts  of  the 
warp  yarn  if  the  counts  in  the  cloth  are  found  to  be  3.15-run? 


Solution. —  20  per  cent.  —  10  per  cent.  =  10  per  cent. 


3.15  X  100 
100  -  10 


3.5-run.  Ans. 


Note.— If  in  this  example  the  shrinkage  in  finishing  and  the  contraction  in 
weaving  were  considered  separately,  the  example  would  have  to  be  solved  in  two 
steps,  first  finding  the  counts  of  the  yarn  in  the  woven  cloth  and  then  the  counts  of 
the  yarn  on  the  beam. 

Example  2. — A  piece  of  cheviot-finish  worsted  cloth  has  shrunk 
18  per  cent,  in  width  from  the  reed  to  the  finished  cloth.  If  the 
counts  of  the  filling  in  the  finished  cloth  are  31.68s  and  the  cloth  has 
lost  6  per  cent,  during  finishing,  what  were  the  original  counts  of  the 
filling? 

Solution. —  18  per  cent.  —  6  per  cent.  =  12  per  cent. 


31.68  X  100 
100  -  12 


36s.  Ans. 


In  a  manner  similar  to  that  explained  for  finding  the 
original  counts  of  the  yarn,  the  weight  of  the  cloth  from  the 


14 


CLOTH  CALCULATIONS, 


13 


loom  may  be  found  if  the  finished  weight,  the  shrinkage  in 
length,  and  the  loss  in  finishing  are  known.  The  shrinkage 
of  the  cloth  in  width  must  be  taken  into  consideration  in  cal¬ 
culations  for  finding  the  width  of  the  cloth  in  the  reed,  etc. 


FANCY  PATTERNS 

19.  When  it  is  desired  to  find  the  number  of  ends  of  each 
color  or  count  in  the  warp  from  a  piece  of  cloth  containing  a 
warp  pattern,  the  following  method  may  be  used  to  advan¬ 
tage  in  some  instances,  especially  if  the  cloth  is  a  stripe 
pattern. 

20.  To  find  the  number  of  ends  of  each  count  or  color  m 
a  cloth  when  different  counts,  colors,  or  materials  are  used: 


Rule. — Find  the  number  of  patterns  in  the  cloth  by  dividing 
the  total  width  by  the  width  of  one  pattern  and  multiply  the 
result  thus  obtained  by  the  number  of  ends  of  each  color  in  the 
pattern.  The  result  will  in  each  instance  be  the  number  of  ends 
of  that  particular  counts  or  color  in  the  warp. 

Example. — A  small  sample  of  worsted  cloth  is  found  to  be  dressed 
12  black,  2  slate,  12  black,  and  2  white,  and  it  is  desired  to  make  a 
similar  cloth  56  inches  wide;  how  many  ends  of  each  color  will  there 
be  in  the  warp? 

Solution. — With  a  small  steel  rule  and  a  pair  of  dividers  it  is  found 
that  there  are  exactly  three  patterns  in  1  inch  of  the  sample.  Then, 
one  pattern  occupies  i  inch.  In  each  pattern  there  are  24  ends  of 
black,  2  ends  of  slate,  and  2  ends  of  white. 


56  -p  §■  =  168  patterns  in  the  cloth 
168  X  24  =  4,032  ends  of  black 


168  X  2  = 
168  X  2  = 


336  ends  of  slate 
336  ends  of  white 


Ans. 


This  gives  a  total  of  4,032  +  336  +  336=4,704  ends  in  the 
warp,  but  as  in  this  example  no  allowance  has  been  made 
for  selvages,  it  will  be  assumed  that  36  ends  are  drawn 
on  each  side  of  the  warp  for  a  selvage,  or  72  ends  for  both 
selvages.  The  total  number  of  ends  in  the  above  warp  will 
then  be  4,704  +  72  =  4,776  ends. 


13 


WOOLEN  AND  WORSTED 


15 


Continuing  still  further  with  this  same  example,  it  will  be 
supposed  that  it  is  desired  to  find  the  reed  to  use  and  the  width 
that  this  cloth  must  be  set  in  the  loom  to  finish  to  the  desired 
width.  Suppose  that  this  particular  kind  of  cloth  shrinks 
15  per  cent,  from  reed  to  finished  cloth;  that  is,  the  total 
shrinkage,  including  the  contraction  from  reed  to  cloth  and 
the  shrinkage  in  finishing,  is  15  per  cent,  in  width.  In  order 
to  find  the  reed  to  use,  the  ends  per  inch  in  the  loom  must 
be  found.  It  must  be  thoroughly  understood  in  finding  this 
item  that,  as  the  cloth  shrinks  15  per  cent,  from  reed  to 
finished  cloth,  the  width  in  the  reed  represents  100  per  cent, 
and  the  width  of  the  finished  cloth  is  85  per  cent,  of  this  width. 

21.  To  find  the  width  in  the  loom  inside  selvages: 


Rule. — Multiply  the  finished  width  inside  of  selvages  by  100 
and  divide  the  result  thus  obtained  by  100  minus  the  percentage 
of  total  shrinkage  from  reed  to  finished  cloth. 

Example. — The  same  as  in  Art.  20. 


Solution. — 
56  X  100 
100-15 


65.88  in.,  width  in  reed  inside  selvages.  Ans. 


22.  To  find  the  number  of  ends  per  inch  in  the  reed: 

Rule. — Divide  the  total  number  of  ends  in  the  warp  exchisive 
of  the  selvage  ends  by  the  width  in  the  reed  inside  of  the  selvages. 

Example. — The  same  as  in  Art.  20. 

Solution. —  4,704  -5-  65.88  =  71.40  ends  per  in.  in  reed.  Ans. 

In  a  case  like  this  it  is  customary  to  take  the  nearest 
number  that  will  divide  evenly,  in  order  to  obtain  a  reed  that 
is  in  stock  if  possible,  so  that  in  this  case  it  will  be  assumed 
that  there  are  72  ends  per  inch  in  the  reed. 

23.  To  find  the  reed  when  the  ends  per  inch  in  the  loom 
are  known: 

Rule. — Divide  the  ends  per  inch  in  the  loom  by  the  mimber  of 
ends  per  de?it  that  are  suitable  for  the  warp  yarn  and  the  fabric 
being  woven. 


16  '  CLOTH  CALCULATIONS,  §13 


Example. — Suppose  that  in  the  sample  considered  in  Art.  22  the 
fabric  is  such  that  it  should  be  reeded  four  in  a  dent;  find  the  reed. 

Solution. —  72  h-  4  =  18s  reed.  Ans. 

In  the  above  fabric,  the  calculated  number  of  ends  per  inch  in 
the  loom,  71.40,  was  not  used,  owing  to  the  reed;  so  in  order 
to  find  the  actual  width  in  the  reed,  it  will  now  be  necessary 
to  use  an  18s  reed,  four  in  a  dent,  or  72  ends  per  inch  in 
the  loom. 


4,704 
4  X  18 


65i  inches,  width  in  loom  inside  of  selvages 


It  was  stated  that  72  ends  of  selvage,  or  36  ends  on  each 
side,  would  be  added  to  this  warp.  These,  if  reeded  six  in 
a  dent  with  an  18s  reed,  will  make  exactly  a  3-inch  selvage 
on  each  side  of  the  cloth;  therefore,  to  find  the  width  of  the 
cloth  in  the  reed  over  selvages,  it  is  simply  necessary  to 
add  f  inch  to  the  width  in  the  reed  inside  selvages;  thus, 
65i  -f-  f  =  66  inches. 

24.  Although  rules  have  been  given  for  finding  the 
counts,  or  size,  of  a  yarn  when  the  weight  of  a  given 
number  of  yards  is  known,  it  is  well  to  give  here  the  rule  for 
finding  the  size  of  the  yarn  under  other  conditions,  such  as 
occur  when  analyzing  a  sample  of  cloth. 


25.  To  find  the  size  of  a  given  yarn  when  the  weight  of 
a  definite  number  of  inches  is  known: 


Rule. — Multiply  the  number  of  inches  weighed  by  7 ,000 
{ grams  in  1  pound )  and  divide  the  result  thus  obtained  by 
the  weight ,  in  grains ,  times  the  standard  number  times  36 
( inches  in  1  yard.) 

Example  1. — It  is  found  that  29  inches  of  woolen  yarn  weighs 
.7  grain;  what  is  the  size  of  the  yarn? 

c  29  X  7.000  _  _0  . 

Solution.  7x  1,600  X  36  =  6  03-run'  Ans' 

Example  2. — It  is  found  that  58  inches  of  worsted  yarn  weighs 
1.4  grains;  what  are  the  counts  of  the  yarn? 

58  X  7,000 
1.4  X  560  X  36 


Solution. — 


=  14.38s.  Ans. 


13 


WOOLEN  AND  WORSTED 


17 


26.  Nothing  has  yet  been  said  of  the  weight  of  cloth, 
which  is  in  itself  an  important  item  and  deserves  the  careful 
attention  of  the  student. 

27.  To  find  the  weight  per  yard  of  finished  cloth,  in 
ounces,  when  the  weight  of  1  square  inch,  in  grains,  is 
known: 

Rule. — Multiply  the  weight  of  1  square  inch ,  in  grains ,  by 
the  width  of  the  cloth  and  by  36  ( inches  i?i  1  yard),  and  divide 
by  437.5  (grains  in  1  ounce). 

Example. — A  square  inch  of  worsted  dress  goods  weighs  2.2  grains. 
If  the  cloth  is  44  inches  wide  when  finished,  what  is  the  weight  per  yard? 

0  2.2  X  44  X  36  „ 

Solution. —  - . -  =  /.96oz.  Ans. 

46/  .5 

28.  To  find  the  weight  of  the  warp  in  1  yard  of  finished 
cloth: 

Rule. — Multiply  the  ends  per  inch  by  the  width  of  the  cloth, 
thus  obtaining  the  ends  in  the  warp,  and  by  16  (ounces  in 
1  pound),  and  divide  the  result  thus  obtained  by  the  size  of  the 
yarn  in  the  finished  cloth  times  its  standard  number. 

Example. — In  the  cloth  mentioned  in  the  previous  example  there  are 
44  warp  threads  per  inch,  and  the  counts  of  the  yarn  in  the  finished  cloth 
are  13.88s;  what  is  the  weight  of  the  warp  in  1  yard  of  finished  cloth? 

„  44  X  44  X  16 

Solution.  i3.88  x  6W  =  3  98  °*'  Ans' 

29.  To  find  the  weight  of  filling  in  1  yard  of  finished 
cloth: 

Rule. — Multiply  the  picks  per  inch  in  the  finished  cloth  by 
the  width  and  by  16  (ounces  in  1  pound),  and  divide  the  result 
thus  obtahied  by  the  counts  of  the  filling  in  the  finished  cloth 
multiplied  by  the  sta?idard  number. 

Example. — Suppose  that  in  the  same  piece  of  cloth  as  in  the 
previous  examples  there  are  41  picks  per  inch  and  the  size  of  the 
filling  in  the  finished  cloth  is  12.94s;  find  the  weight  of  filling  in  1  yard 
of  finished  cloth. 

Solution.-  ~12,94  x  660  =  3.98  os.  Ans. 

In  this  piece  of  cloth  it  will  be  noticed  that  the  weight  of 
warp  and  filling  in  a  yard  of  cloth  is  the  same.  To  find  the 


18 


CLOTH  CALCULATIONS, 


13 


weight  of  1  yard  of  finished  cloth  it  is  now  only  necessary  to 
add  the  weights  of  the  warp  and  filling,  in  this  case  3.98  and 
3.98  ounces,  which,  when  added,  equal  7.96  ounces.  It  will 
be  seen  that  this  is  a  method  of  proving  the  weight  of  1  yard 
of  finished  cloth  as  found  from  the  weight  of  1  square  inch. 

Note. — In  Arts.  27,  28,  and  29  the  width  of  the  finished  cloth  is 
considered  inside  of  the  selvages,  and  the  calculations  and  rules  for¬ 
mulated  on  this  basis.  The  reason  for  this  is  that  when  analyzing  a 
small  sample  of  cloth,  the  required  finished  width  is  usually  given 
inside  of  the  selvages,  and  by  ignoring  the  selvages  at  this  point  the 
weight  per  yard  of  the  cloth,  as  found  from  the  weight  in  grains  of 
1  square  inch,  can  be  readily  proved,  as  explained.  The  desired  num¬ 
ber  of  ends  for  selvages  can  afterwards  be  added  to  the  warp  and 
width  in  reed  in  finding  other  particulars. 

30.  The  weight  of  the  finished  cloth  is  easily  obtained 
by  means  of  these  rules,  but  in  cases  where  it  is  desired  to 
know  the  weight  of  the  cloth  from  the  loom  the  following 
method  is  pursued. 

31.  To  find  the  weight  of  the  warp  yarn  in  a  yard  of 
woven  cloth  from  the  loom: 

Rule. — Multiply  the  number  of  ends  in  the  warp ,  including 
the  selvage  e?ids,  by  16  ( dunces  in  1  pound) ,  a?id  divide  the  result 
thus  obtained  by  the  original  size  of  the  warp  yarn  multiplied 
by  the  standard  number.  To  the  result  tlms  obtained ,  add  the 
percentage  of  take-up  in  weaving. 

Example. — A  woolen  warp  in  a  piece  of  cloth  contains  840  ends  and 
30  selvage  ends;  the  original  size  of  the  yarn  is  2.14-run  and  the  warp 
takes  up  6  per  cent,  in  weaving;  what  is  the  weight  of  warp  in  the  cloth 
from  the  loom? 


Solution. — 


4.06  oz.  +  6  per  cent.  =  4.30  oz.  Ans. 


While  it  is  true  that  the  cloth  takes  up,  or  shrinks,  from 
the  reed  to  the  cloth,  or  in  width,  and  this  is  take-up,  or 
shrinkage,  in  weaving,  it  does  not  affect  the  weight  of  the 
cloth,  as  cloth  is  measured  and  weighed  by  the  linear  yard; 
therefore,  it  is  only  the  shrinkage  of  the  cloth  in  length  that 
affects  its  weight. 

32.  To  find  the  weight  of  filling  in  a  yard  of  woven 
cloth  from  the  loom: 


13 


WOOLEN  AND  WORSTED 


19 


Rule. — Multiply  the  width  of  the  cloth  in  the  reed  in  inches 
by  the  picks  per  inch  and  by  16  ( ounces  in  1  pound)  and  divide 
the  result  thus  obtained  by  the  size  of  the  filling  multiplied 
by  the  standard  number . 


Example. — The  cloth  mentioned  in  the  example  in  Art.  31  is 
34.15  inches  in  the  reed  and  contains  27  picks  per  inch  woven;  the  size 
of  the  filling  is  2.14-run;  what  is  the  weight  of  filling  in  1  yard  of 
cloth  from  the  loom? 


Solution. — 


34.15  X  27  X  16 
2.14  X  1,600 


4.30  oz.  Ans. 


33.  To  find  the  weight  of  the  cloth  from  the  loom  it  is 
simply  necessary  to  add  the  weights  of  the  warp  and  filling, 
which  in  the  above  examples,  it  will  be  noticed,  are  equal. 
Therefore,  the  weight  of  the  above  cloth  from  the  loom  is 
equal  to  4.30  ounces  +  4.30  ounces,  or  8.60  ounces. 


EXAMPLES  FOR  PRACTICE  ,7  ^ 

1.  A  warp  contains  2,924  ends  and  is  dressed  as  follows:  8  black, 
2  slate,  12  black,  4  slate,  4  black,  3  slate,  and  1  fancy;  how  many  ends 
of  each  color  are  there  in  the  warp?  r 2,064  black 

Ans.  <  774  slate 
186  fancy 


2.  A  warp  containing  1,800  ends  is  drawn  in  according  to  the  draft, 
shown  in  Fig.  1;  how  many  heddles  are  required  for  each  harness? 

’540  heddles  on  first  harness 
A  720  heddles  on  second  harness 
'  360  heddles  on  third  harness 
180  heddles  on  fourth  harness 


7 

7 

7 

6 

6 

6 

6 

5 

5 

5 

5 

4 

4 

4 

4 

3 

3 

3 

3 

2 

2 

2 

2 

1 

1 

1 

Fig.  2 


3.  A  warp  containing  3,900  ends  is  drawn  in  according  to  the  draft 
shown  in  Fig.  2;  how  many  heddles  will  be  required  on  each  harness? 

!450  heddles  on  first  and  seventh  harnesses 
600  heddles  on  second,  third,  fourth,  fifth, 
and  sixth  harnesses 


20 


CLOTH  CALCULATIONS, 


13 


4.  A  woolen  warp  is  drawn  three  per  dent  in  an  8s  reed  and  con¬ 
tains  912  ends;  how  much  reed  space  does  the  warp  occupy? 

Ans.  38  in. 

5.  A  worsted  warp  is  reeded  four  per  dent  in  a  15s  reed  and  set 
34  inches  wide;  how  many  ends  are  there  in  the  warp? 

Ans.  2,040  ends 

6.  A  warp  containing  2,700  ends  is  set  30  inches  wide  in  the  reed 
and  is  reeded  five  per  dent;  what  is  the  number  of  the  reed  used? 

Ans.  18s  reed 

7.  A  finished  cloth  is  28  inches  wide  inside  of  selvages  and  has 
shrunk  12  per  cent,  from  reed  to  finished  cloth;  if  the  selvage  occupies 
1  inch  in  the  reed,  what  is  the  total  width  of  the  cloth  in  the  loom? 

Ans.  32.81  in. 

8.  How  many  yards  of  cloth  from  the  loom  will  be  obtained  from 
240  yards  of  warp,  if  the  contraction  in  weaving  is  8  per  cent.? 

Ans.  220.8  yd. 

9.  A  warp  contains  2,592  ends  and  is  reeded  four  per  dent  in  an 

18s  reed;  how  wide  is  the  cloth  in  the  loom?  Ans.  36  in. 

10.  A  finished  piece  of  cloth  is  60  yards  in  length  and  has  shrunk 
20  per  cent,  in  finishing;  how  long  was  the  unfinished  piece? 

Ans.  75  yd. 


§  13 


WOOLEN  AND  WORSTED 


21 


FIGURING  PARTICULARS  FROM  CLOTH 

SAMPLES 

34.  In  order  that  the  method  of  analyzing’  a  sample  of 
cloth  may  be  understood,  a  sample  of  cloth  is  enclosed  and 
the  method  of  procedure  will  be  explained  with  reference  to 
this  sample.  It  is  not  intended  at  this  stage  to  teach  the 
method  of  picking  out  the  weave  and  of  obtaining  the  draw¬ 
ing-in  and  chain  drafts,  but  simply  to  acquaint  the  student 
with  the  method  of  figuring  the  given  sample  with  a  view  to 
its  reproduction. 

The  sample  of  cloth  that  has  been  taken  to  illustrate  the 
process  of  cloth  analysis  is  a  piece  of  worsted  dress  goods 
finished  48  inches  wide  inside  the  selvages. 


WEIGHT  OF  CLOTH 

35.  The  first  operation  in  analyzing  a  sample  of  cloth 
is  to  find  the  weight,  in  grains,  of  1  square  inch  of  the 
cloth.  This  is  usually  accomplished  by  cutting  out  a  square 
inch  with  a  steel  die  and  weighing  it  with  a  grain  scale. 
A  steel  rule  and  a  sharp  knife  are  also  often  used  for  cut¬ 
ting  out  a  given  area  of  cloth.  Quite  often  2  or  3  square 
inches  are  cut  if  the  sample  of  cloth  is  large  enough  to 
allow  it,  and  the  resulting  weight  divided  by  2  or  3,  as  the 
case  may  be.  In  the  cloth  under  consideration,  it  has  been 
found  that  6  square  inches  of  cloth  weigh  exactly  8  grains. 
The  first  calculation  is  to  find  the  weight  of  the  cloth  in 
ounces  per  yard  from  the  data  that  are  now  possessed  con¬ 
cerning  this  sample,  namely,  the  weight  of  6  square  inches 

and  the  finished  width.  =  5-26  ounces,  finished 

6  X  437.5 

weight  exclusive  of  selvages. 


22 


CLOTH  CALCULATIONS, 


§13 


COUNTS  OF  THE  WARP  AND  FILLING 


36.  Next,  it  is  desired  to  find  the  size,  or  counts,  of  the 
yarns  in  this  sample  of  cloth.  It  is  found  that  the  brown 
and  the  white  warp  yarns  are  exactly  the  same  size,  and 
that  100  inches  of  the  warp  yarn  weighs  exactly  1  grain. 
Then,  as  this  is  worsted  yarn,  the  standard  of  which  is 
560  yards  to  the  pound,  this  number  must  be  used  in  finding 

the  counts  of  the  yarn. - — ^ — —  =  34. /2s,  counts  of 

1  X  560  X  36 

warp  yarn. 

Proceeding  in  the  same  manner,  it  is  ascertained  that  the 
counts  of  the  brown  and  white  filling  are  the  same,  and  that 
104  inches  of  filling  yarn  is  required  to  weigh  1  grain. 


104  X  7,000 
1  X  560  X  36 


36.11s,  counts  of  filling  yarn. 


Ans. 


NUMBER  OF  ENDS  IN  THE  WARP 

37.  It  is  next  desired  to  find  the  number  of  ends  in  the 
warp.  By  counting  with  a  pick  glass  or  by  measuring  off 
a  given  space  and  counting  the  ends  it  is  found  that  there 
are  74  ends  per  inch,  and,  as  the  cloth  is  48  inches  wide, 
it  would  naturally  be  assumed  that  there  are  48  X  74,  or 
3,552  ends,  in  the  warp.  Before  the  number  of  ends  in  the 
entire  warp  is  finally  determined  on,  however,  it  is  best  to 
see  if  the  number  of  ends  in  the  warp  is  divisible  by  the 
number  of  ends  in  the  pattern,  and,  if  not,  to  increase  or 
decrease  the  ends  in  the  warp,  as  the  case  may  be,  in  order 
to  make  even  patterns.  In  this  case  it  is  found  that  there 
are  24  ends  in  one  pattern  of  the  warp;  there  are  therefore 
148  repeats  in  the  total  width  of  48  inches. 


PROOF  OF  WEIGHT  AND  COUNTS 

38.  Having  found  the  number  of  ends  in  the  warp 
and  the  size  of  the  yarns  used,  the  next  operation  will 
be  to  find  the  weight  of  the  cloth,  as  figured  from  the 
yarn,  in  order  to  prove  that  these  items  have  been  found 
correctly. 


§13 


WOOLEN  AND  WORSTED 


23 


39.  Weight  of  the  Warp. — To  accomplish  this  it  is 
first  necessary  to  find  the  weight  of  the  warp  per  yard  of 


cloth. 


3,552  X  16 
34.72  X  560 


2.92  ounces,  weight  of  warp  yarn. 


Had  there  been  two  or  more  counts  of  yarn  in  the  warp  in 
this  sample  of  cloth,  it  would  have  been  necessary  to  find 
the  number  of  ends  and  weight  of  each  and  add  the  results, 
or  to  find  the  average  counts  of  the  warp  yarn,  whichever 
would  have  been  most  convenient. 


40.  Weight  of  the  Filling. — It  is  next  necessary  to 
find  the  weight  of  the  filling  per  yard  of  cloth  and,  in  order 
to  accomplish  this,  the  number  of  picks  per  inch  must  be 
determined.  This  can  be  done  in  the  same  manner  as  when 
finding  the  ends  per  inch.  The  sample  under  considera¬ 
tion  will  be  found  to  have  62  picks  per  inch. 


48  X  62  X  16 
36.11  X  560 


2.35  ounces,  weight  of  filling  yarn. 


If  the  filling  in  the  sample  under  consideration  had  been 
composed  of  two  or  more  different  counts  of  yarn,  it  would 
have  been  necessary  to  find  either  the  weight  of  each  or 
the  average  counts  of  the  filling. 

The  weight  of  the  warp  as  previously  found  was  2.92  ounces, 
which,  when  added  to  2.35  ounces,  the  weight  of  the  filling, 
equals  5.27  ounces,  which  is  the  weight  of  1  yard  of  the 
finished  cloth,  exclusive  of  the  selvage.  It  will  be  noticed 
that  this  result  proves  within  .01  the  weight  of  1  yard  as  found 
from  the  weight  of  6  square  inches;  therefore,  the  sizes  of 
the  yarn  and  weights  of  warp  and  filling  must  be  correct.  It 
is  not  always  possible  to  prove  the  work  as  close  as  this, 
but  it  is  advisable  to  prove  as  close  as  .05,  which  allows  for 
small  errors  occurring  through  not  carrying  out  the  decimals. 
If  the  error  is  greater  than  .05,  it  is  better  to  go  over  the 
work  again,  although  some  designers  only  prove  within  .1. 


24 


CLOTH  CALCULATIONS, 


13 


REED  AND  ENDS  PER  DENT 


41.  The  next  item  of  importance  to  decide  on  with 
regard  to  this  piece  of  cloth  is  the  reed,  in  order  to  find 
which  it  is  necessary  to  find  the  ends  per  inch  in  the  reed. 
This  being  a  clear-finished  worsted,  the  total  shrinkage  from 
the  reed  to  the  finished  cloth  is  about  10  per  cent. 


In  a  case  like  this  it  is  customary  to  drop  the  fraction  and 
make  a  whole  number  of  ends  per  inch  in  the  loom,  in  order 
that  a  reed  may  be  readily  obtained;  in  this  case  say 
66  ends  per  inch  in  the  reed,  drawn  3  ends  in  1  dent. 


66  -4-  3  =  22s  reed.  Ans. 


WIDTH  IN  REED 


42.  It  is  next  desired  to  find  the  width  of  the  cloth  in  the 
reed.  3,552  4-  66  =  53.81  inches  in  the  reed.  This  width  in 
the  reed  is  of  course  inside  of  the  selvages. 

Suppose  that  in  this  fabric  7  dents  are  used  for  each  sel¬ 
vage,  or  14  dents  for  both  selvages.  Then  the  width  of  the 
selvages  in  the  reed  will  equal  14  divided  by  22,  or  .63  inch, 
and  the  total  width  of  the  fabric  in  the  reed  will  be  53.81  inches 
plus  .63  inch,  or  54.44  inches. 


WEIGHT  OF  CLOTH  FROM  LOOM 


43.  It  is  next  desired  to  find  the  weight  of  the  woven  cloth 
from  the  loom.  If  6  ends  per  dent  are  drawn  for  selvages 
of  the  same  yarn  as  the  warp,  there  will  be  14  X  6  =  84  sel¬ 
vage  ends,  which,  added  to  the  ends  in  the  body  of  the  cloth, 
will  make  3,636  ends  in  the  whole  warp.  The  sample  of 
cloth  being  a  clear-finished  worsted,  it  will  be  assumed  that 
the  shrinkage  from  warp  yarn  to  finished  cloth  and  from  reed 
to  finished  cloth  is  10  per  cent.,  and  that  the  yarn  will  lose 
10  per  cent,  in  finishing,  so  that  the  original  counts  of  the 
yarn  in  warp  and  filling  will  be  the  same  as  found  in  the 
finished  cloth. 


§13 


WOOLEN  AND  WORSTED 


25 


Had  this  cloth  been  woolen  and  shrunk  very  much,  it  would 
have  been  necessary  to  find  the  original  size  of  the  warp  and 
filling  yarns,  but  generally  speaking,  in  worsted  goods,  the 
loss  in  finishing  will  counterbalance  the  gain  in  weight  of 
the  yarn  due  to  the  shrinkage. 


44.  Weiglit  of  Warp. — The  first  step  in  finding  the 
weight  of  the  cloth  from  the  loom  is  to  find  the  weight  of 
1  yard  of  the  warp  yarn,  including  the  selvage  yarn. 

3,636  X  16  _  2.99  ounces,  weight  of  1  yard  of  warp.  As  this 
34.72  X  560  ^ 

cloth  will  take  up  in  weaving  about  4  per  cent.,  it  is  necessary  to 
add  this  to  the  weight  of  the  warp,  in  order  to  find  the  weight  of 
the  warp  in  1  yard  of  the  cloth  from  the  loom .  2.99  +  4  per  cent. 
=  3.10  ounces,  weight  of  warp  in  1  yard  of  cloth  from  loom. 


45.  Weight  of  Filling. — It  is  next  necessary  to  find  the 
weight  of  the  filling  from  the  loom,  in  order  to  find  the 
weight  of  the  cloth  from  the  loom.  It  was  found  that  there 
were  62  picks  per  inch  in  the  finished  cloth,  but  as  the  finished 
cloth  has  undergone  a  shrinkage  of  10  per  cent,  in  the  length, 
the  number  of  picks  must  have  been  increased  10  per  cent.,  so 
that  62  picks  represent  110  per  cent,  of  the  original  number  of 


picks  per  inch. 


62  X  100 


=  56.36.  This  number  represents 


100  +  10 

the  picks  put  in  by  the  loom,  but  as  the  warp  takes  up 
4  per  cent,  in  weaving,  and  contracts  after  being  taken  from 
the  loom  there  must  be  4  per  cent,  more  picks  in  the  cloth 
from  the  loom,  which  must  be  taken  into  account  in  finding 
the  weight  of  the  cloth  from  the  loom.  56.36  +  4  per 
cent.  =  58.61  picks  per  inch  in  cloth.  Having  found  the 
picks  per  inch  in  the  woven  cloth  and  knowing  the  width 
in  the  loom,  the  weight  of  the  filling  is  easily  obtained. 
54.44  X  58.61  X  16 


36.11  X  560 


2.52  ounces,  weight  of  filling  in  1  yard 


of  cloth  from  the  loom. 

Adding  the  weight  of  warp  and  filling,  the  weight  of  the 
cloth  from  the  loom  is  obtained.  3.10  +  2.52  =p  5.62  ounces 
from  loom. 


26 


CLOTH  CALCULATIONS, 


§13 


SUMMARY 

The  data  obtained  from  the  sample  of  cloth  and  the  entire 
figuring  for  this  sample  of  cloth  are  as  follows: 


Data 

6  square  inches  weighs  8  grains. 

100  inches  of  warp  yarn  weighs  1  grain. 

104  inches  of  filling  yarn  weighs  1  grain. 

10-per-cent,  shrinkage  from  reed  to  finished  cloth. 
10-per-cent,  shrinkage  from  warp  yarn  to  finished  cloth. 
10-per-cent,  loss  of  weight  in  finishing. 

4-per-cent,  take-up  of  warp  in  weaving. 

48  inches,  finished  width  inside  of  selvages. 

74  ends  per  inch,  finished. 

62  picks  per  inch,  finished. 

24  ends  in  a  pattern  (warp  and  filling). 


Analysis 


48  X  36  X  8  = 
6  X  437.5 
100  X  7,000 
1  X  560  X  36 
104  X  7,000 
1  X  560  X  36 


5.26  ounces,  finished  weight,  inside  selvages. 
=  34.72s,  counts  of  warp  yarn. 

=  36.11s,  counts  of  filling  yarn. 


74  X  48  =  3,552  ends  in  warp. 

3,552  -i-  24  =  148  patterns  in  warp. 


3,552  X  16 
34.72  X  560  ' 
48  X  62  X  16 
36.11  X  560 


-  2.92  ounces,  weight  of  warp  in  finished  cloth. 
=  2.35  ounces,  weight  of  filling  in  finished 


cloth. 

2.92  ounces  +  2.35  ounces  =  5.27  ounces,  proof  of  finished 
weight  as  found  from  the  weight  of  6  square  inches. 


3,552  -r- 


/ 48  X  100\ 

Vioo  - 10/ 


66 +,  say  66  ends  per  inch  in  reed. 


66  -h  3  =  22s  reed. 


3,552  -j-  66  =  53.81  inches*  width  in  reed  inside  selvages. 


13 


WOOLEN  AND  WORSTED 


27 


14  -f-  22  =  .63  inch  for  selvages. 

53.81  +  .63  =  54.44  inches,  total  width  in  reed. 

14  X  6  =  84  ends  for  selvages. 

3,552  +  84  =  3,636,  total  ends  in  warp. 

3,636  X  16  0  on  •  u*.  £  i  ^  t 

-  nn - =  2.99  ounces,  weight  of  1  yard  of  warp. 

34.72  X  560 

2.99  ounces  +  4  per  cent.  =  3.10  ounces,  weight  of  warp 
in  1  yard  of  woven  cloth. 

g*™  _  5g  30.  50,30  _f_  4  per  cent.  =  58.61  picks  in  the 


cloth  from  loom. 


54,44  X  58. -61  X  16 
36.11  X  560 


=  2.52  ounces,  weight  of  filling  in 


1  yard  of  cloth  from  loom. 

3.10  ounces  +  2.52  ounces  =  5.62  ounces,  weight  of  cloth 
from  loom. 


It  is  always  best  to  figure  a  sample  through  carefully,  using 
the  exact  counts  of  yarn  as  found  in  the  cloth  and  other  items, 
in  general,  of  the  exact  value  found,  and  afterwards  make 
any  allowances  that  may  be  deemed  necessary  to  facilitate 
the  process  of  manufacture.  For  instance,  if  it  were  desired 
to  reproduce  this  sample,  it  would  be  best  to  spin  36s  yarn  for 
both  warp  and  filling,  as  it  would  be  inadvisable  to  spin  34s 
for  the  warp  and  36s  for  the  filling.  It  mignt  also  be  neces¬ 
sary  to  alter  the  picks  per  inch,  since  it  might  be  found  that 
60  picks  would  produce  a  better  looking  fabric,  etc. 


EXAMPLES  FOR  PRACTICE 

1.  What  weight  of  number  20s  worsted  yarn  will  be  required  for 

filling  in  a  piece  of  cloth  35  inches  wide  at  the  reed,  50  yards  long,  and 
containing  60  picks  per  inch?  Ans.  9-f  lb. 

2.  If  44  inches  of  worsted  yarn  weighs  .7  grain,  what  is  the  size  of 

the  yarn?  Ans.  21.82s  yarn 

3.  What  is  the  weight  of  1  yard  of  cloth  35  inches  wide  if  1  square 

inch  weighs  1.7  grains?  Ans.  4.89  oz. 

4.  What  is  the  weight  of  1  yard  of  warp  containing  1,008  ends 

of  3-run  yarn?  Ans.  3.36  oz. 


28 


CLOTH  CALCULATIONS 


13 


5.  The  warp  yarn  in  a  piece  of  cloth  is  found  to  be  17s;  if  this 
cloth  has  shrunk  25  per  cent,  in  length  and  lost  10  per  cent,  in  weight 
in  finishing,  what  was  the  original  number  of  the  yarn?  Ans.  20s 

6.  A  certain  cloth  is  made  from  3-run  yarn,  both  warp  and  filling. 

The  warp  contains  1,044  ends,  including  the  selvages,  and  is  reeded 
35  inches  wide  in  the  loom.  There  are  29  picks  per  inch  in  the  cloth 
from  the  loom.  The  take-up  of  the  warp  in  weaving  is  4  per  cent. 
What  is  the  weight  of  this  cloth  from  the  loom?  Ans.  7  oz. 

7.  One  square  inch  of  a  28-inch  cloth  weighs  2.75  grains;  what  is 

the  weight  of  the  cloth  in  ounces  per  yard?  Ang.  6.33  oz. 

8.  If  40  inches  of  woolen  yarn  weighs  1.2  grains,  what  run  is 

the  yarn?  Ans.  4.05-run 

9.  A  certain  warp  contains  1,920  ends  of  2/40s  worsted;  how  many 

pounds  of  warp  will  it  take  to  weave  a  50-yard  cut  of  cloth  if  the  warp 
contraction  in  weaving  is  6  per  cent.?  Ans.  9.118  lb. 

10.  What  is  the  weight  of  filling  in  1  yard  of  cloth  32  inches  wide 

containing  30  picks  per  inch  of  4-run  woolen  yarn?  Ans.  2.4  oz. 


DRAFT  CALCULATIONS 


DRAFTING 


INTRODUCTION 

1.  In  the  manufacture  of  cotton  yarn  a  principle  is 
adopted  that  has  to  be  considered  in  connection  with  almost 
every  process  between  the  opening-  of  the  raw  cotton  and  the 
spinning  of  the  yarn  — that  known  as  drafting.  This  prin¬ 
ciple  is  one  of  great  importance  to  any  one  desiring  to 
obtain  a  thorough  knowledge  of  cotton-yarn-mill  machinery, 
and  should  be  the  subject  of  careful  study  and  constant 
watchfulness  in  order  that  the  best  results  may  be  obtained. 
As  the  principle  of  drafting  cannot  be  considered  a  special 
feature  of  any  one  machine,  and  as  it  must  be  considered  so 
frequently  in  connection  with  many  machines,  a  general 
explanation  of  it  is  here  given. 

The  word  drafting  in  the  cotton-mill  business  is  applied 
to  the  principle  of  attenuating,  or  drawing  out,  a  compara¬ 
tively  large  mass  of  cotton  fibers  to  the  condition  of  a 
thinner  but  longer  mass.  The  fibers  found  in  the  cotton 
bale  possess  no  regular  order  or  arrangement.  The 
machines  through  which  the  cotton  passes  in  the  early 
stages  of  its  manufacture  into  cotton  yarn  arrange  these 
fibers  into  some  sort  of  order  until  they  assume  the  form  of 
the  sliver ,  which  is  a  rope  of  loose  cotton  in  which  the  fibers 
lie  side  by  side,  or  as  nearly  so  as  it  is  possible  to  arrange 
them.  Some  fibers  overlap,  while  others  are  end  to  end,  but 
the  object  is  to  arrange  them  all  approximately  parallel  so 


For  notice  of  copyright,  see  page  immediately  following  the  title  page 

§15 


2 


DRAFT  CALCULATIONS 


15 


that  the  ribbon  or  rope  of  fibers  can  be  handled  without  being 
broken,  and  so  that  the  weight  of  1  yard  of  the  sliver  may 
be  approximately  the  same  as  that  of  another  yard,  varying 
only  to  the  extent  of  a  few  grains. 

Previous  to  this  form,  the  cotton  has  been  arranged  in  a 
sheet  36  or  40  inches  wide,  and  made  into  a  roll,  but  without 
any  attempt  having  been  made  at  parallelizing  the  fibers;  in 
this  condition  it  is  known  as  a  lap.  The  expression  draft 
is  applied  to  indicate  the  extent  to  which  these  laps  are 
drawn  out  or  elongated  to  form  other  laps;  to  the  resultant 
lap  drawn  out  to  make  a  sliver;  to  the  slivers  drawn  out  to 
make  still  smaller  slivers,  known  as  rovings ;  or  to  the  extent 
to  which  these  rovings  are  drawn  out  to  make  still  smaller 
rovings,  or  to  make  yarn. 

The  principle  of  attenuating  the  mass  of  cotton  fibers  by 
various  means  is  one  of  the  fundamental  principles  of  cotton- 
yarn  preparation.  It  may  be  done  by  means  of  air-currents, 
by  which  the  fibers  are  separated  one  from  the  other  and 
carried  along  by  a  current  of  air  and  deposited  on  rotating 
screens  delivering  the  sheet  of  cotton  at  a  higher  speed  than 
that  at  which  it  is  fed  into  the  machine;  it  may  be  performed 
by  rapidly  rotating  cylinders  and  rolls  covered  with  wire 
teeth,  which  elongate  the  mass  of  fibers  even  to  the  extent 
of  separation,  depositing  them  again  at  a  given  rate  on  a 
condenser,  or  doffer;  or  it  may  be,  and  most  frequently  is, 
performed  by  means  of  revolving  rolls.  It  is  to  the  prin¬ 
ciples  of  drafting  by  means  of  successive  pairs  of  revolving 
rolls  that  most  frequent  reference  will  be  made. 

2.  Objects  of  Drafting.— In  attenuating,  or  drawing 
out,  the  mass  of  cotton,  there  are  three  principal  objects: 
the  first  is  to  reduce  the  lap,  sliver,  or  roving  to  a  less  weight 
per  yard,  that  is,  attenuating  it  gradually  to  the  desired 
degree  of  fineness;  the  second  object  is  that  of  arranging 
and  improving  the  arrangement  of  the  fibers  in  a  parallel 
order  so  that  they  may  lie  side  by  side  and  overlap  one 
another;  the  third  object  is  that  of  evening  the  strand  of 
fibers  to  eliminate  thick  or  thin  places,  which  is  done  by  a 


§15 


DRAFT  CALCULATIONS 


3 


combination  of  drafting  and  doubling.  The  use  of  succes¬ 
sive  pairs  of  drawing  rolls  is  largely  adopted  to  arrive  at 
these  results. 

The  action  of  the  drawing  rolls  can  be  imitated  in  the  fol¬ 
lowing  manner:  Grasping  in  the  left  hand  a  tuft  of  loose 
cotton  that  has  been  partially  cleansed  at  one  of  the  earlier 
processes,  with  the  right  hand  pull  the  extreme  projecting 
ends  forwards;  then,  after  obtaining  a  fresh  grip  farther  back 
with  the  left  hand,  and  a  fresh  grip  farther  back  with  the 
right  hand,  pull  still  more  fibers  forwards.  By  repeating 
this,  what  was  a  loose  tuft  of  cotton  will  gradually  be  drawn 
out  to  a  thin  film  of  partially  straightened  fibers. 

This  crude  method  illustrates  that  the  process  of  drawing 
a  number  of  cotton  fibers  past  one  another,  always  holding  the 
front  ends  firmly  and  the  rear  ends  of  the  same  fibers  loosely, 
results  in  the  whole  mass  of  fibers  being  gradually  straight¬ 
ened  out.  This  principle  is  made  use  of  in  most  cotton-yarn- 
preparation  machines  by  having  carefully  constructed  and 
adjusted  rolls,  the  rear  ones  holding  the  mass  of  fibers  and 
running  at  a  slow  speed,  the  forward  ones  tightly  gripping  a 
portion  of  the  fibers  and  revolving  at  a  greater  speed.  This 
arrangement  is  duplicated  again  and  again,  until  in  some 
machines  there  are  as  many  as  four  pair  of  rolls  successively 
acting  on  the  fibers  in  this  manner.  The  quickly  rotating 
pair  of  rolls  draws  the  fibers  away  from  the  slowly  rotating 
rolls,  and  as  the  fibers  are  gripped  by  their  fore  ends  and 
pulled  forwards,  the  loose  rear  ends  trail  behind  and  tend  to 
become  straightened  out  as  they  are  drawn  from  the  tuft  held 
by  the  slowly  rotating  rolls. 

In  early  days,  when  cotton  yarns  were  made  by  hand,  the 
deftness  of  the  spinster’s  fingers  was  relied  on  to  obtain 
this  parallelizing  of  fibers,  but  over  100  years  ago  the 
principle  of  drafting  by  means  of  rolls  was  discovered  by  Sir 
Richard  Arkwright,  and  has  been  in  use  ever  since. 

3.  Doubling. — The  explanation  just  given  shows  that 
attenuating  and  parallelizing  the  mass  of  fibers  tends  to 
reduce  its  thickness  and  make  a  thin  sheet  where  there  was 


4 


DRAFT  CALCULATIONS 


15 


formerly  a  thick  one,  and  if  continued  indefinitely  would 
result  in  destroying  the  continuity  of  the  sliver  or  roving. 
To  prevent  this,  doubling  is  resorted  to  in  most  of  the  cotton- 

yarn-preparation  machines.  Briefly  explained,  this  means 

> 

that  instead  of  feeding  only  one  lap,  sliver,  or  roving  at  the 
back  of  each  machine,  two  or  more  are  fed  together,  making 
one  at  the  front;  this  not  only  helps  to  compensate  for  the 
excessive  attenuation,  but  has  the  great  advantage  of  helping 
to  neutralize  unevenness  in  the  original  mass  of  fiber  fed  to 
the  machine.  By  feeding  several  together,  the  thick  or  thin 
places  of  any  one  are  combined  with  other  slivers  of  normal 
size,  or  thick  places  with  thin  ones,  and  the  combination  of 
two,  three,  four,  five,  or  six  independent  slivers  or  rovings, 
which  are  drawn  out  into  one,  results  in  an  evenness  not 
attainable  in  any  other  manner. 

4.  The  general  operation  or  application  of  these  principles 
is  spoken  of  as  attenuation,  or  drawing  out,  and  sometimes 
as  drafting ,  but  strictly  speaking,  a  more  limited  meaning 
attaches  to  the  word  drafting.  Draft  as  applied  to  cotton- 
yarn  machinery  should  only  be  considered  as  referring  to  the 
ratio  of  attenuation,  and  drafting  should  refer  to  the  attenu¬ 
ation  only,  without  being  understood  to  include  the  parallel¬ 
izing  or  evening  features  that  have  just  been  referred  to, 
although  these  follow  as  a  matter  of  course.  A  full  consid¬ 
eration  of  all  these  features  cannot  be  made  until  the  con¬ 
struction  and  operation  of  the  machines  have  been  dealt  with 
in  their  proper  places,  but  the  subject  of  draft,  pure  and 
simple,  can  now  be  considered  and  the  calculations  connected 
therewith  explained,  so  that  this  information  can  be  applied 
to  each  machine  in  which  draft  rolls  are  used. 

The  word  draft  as  used  here  must  always  be  considered  as 
referring  to  the  ratio  of  attenuation,  and  must  not  be  con¬ 
fused  with  the  word  draft  that  indicates  a  draft  of  air,  as 
used  in  connection  with  the  pneumatic  drafts  found  in  some 
cotton-yarn-preparation  machines. 


15 


DRAFT  CALCULATIONS 


5 


DRAFTING  WITH  COMMON  ROLLS 

5.  The  study  of  draft  calculations  is  of  great  value  for 
many  reasons,  and  its  importance  cannot  be  exaggerated. 
A  thorough  knowledge  of  this  subject  enables  the  amount  of 
draft,  or  attenuation,  that  is  taking  place  in  each  machine  to 
be  determined;  it  enables  the  right  size  of  lap,  sliver,  roving, 
or  yarn  to  be  produced,  or  changes  to  be  made  from  one  size 
to  another;  it  provides  a  means  of  calculating  how  errors  of 
size  in  the  resulting  product  of  any  machine  can  be  corrected, 
and  to  anticipate  beforehand  what  resultant  size  of  lap, 
sliver,  roving,  or  yarn  can  be  made  from  a  machine  or  a 
series  of  machines. 

The  subject  will  be  dealt  with  here  only  so  far  as  the  cal¬ 
culations  are  concerned,  but  before  considering  these  calcu¬ 
lations  a  brief  description  of  the  passage  of  cotton  through 
drafting  rolls,  according  to  one  method,  will  be  given. 
Fig.  1  represents  a  section  though  four  pair  of  rolls,  the 


Fig.  i 


lower  rolls  a ,  b ,  c ,  d  being  constructed  of  steel  and  fluted 
longitudinally.  These  flutes  are  cut  in  such  a  manner  that 
the  outer  edge  of  the  projections  ends  in  almost  a  flat  surface; 
strictly  speaking,  they  are  arcs  of  a  circle  which  has  the  center 
of  the  roll  as  a  center.  The  upper  rolls  alt  blt  f,,  dy  are  con¬ 
structed  of  iron  with  a  covering  of  flannel  immediately 
around  them,  and  a  thin  leather  covering  outside  of  the 
flannel.  These  rolls  are  absolutely  smooth,  and  are  pressed 
against  the  bottom  rolls  by  means  of  weights. 


6 


DRAFT  CALCULATIONS 


§15 


The  rolls  d,  d1  into  which  the  material  is  fed  should  always 
be  spoken  of  a.s  the  feed-rolls  or  back  rolls ,  the  roll  dx  being 
distinguished  from  the  roll  d  by  the  term  back  top  roll.  The 
rolls  delivering  the  material,  represented  by  a  and  aly  should 
always  be  spoken  of  as  the  delivery  rolls  or  front  rolls ,  the 
roll  a ,  being  called  the  front  top  roll.  The  first  pair  of  inter¬ 
mediate  rolls,  counting  from  the  front,  is  spoken  of  as  the 
second  pair  of  rolls;  and  the  third  pair,  counting  from  the 
front,  as  the  third  pair  of  rolls.  Thus,  the  roll  a  would  be 
the  front,  or  delivery,  roll;  b  would  be  the  second  roll;  c,  the 
third  roll;  and  d,  the  back  roll,  or  feed-roll. 

The  circumferential  speed  of  the  upper  and  lower  roll  in 
each  pair  should  be  the  same;  that  is,  a  point  on  the  surface 
of  d  should  move  at  the  same  speed  as  a  point  on  the  sur¬ 
face  of  dly  because  dx  is  driven  by  frictional  contact  with  d. 
The  same  remarks  apply  to  any  other  pair  in  the  series.  In 
practice,  however,  it  is  found  that  there  is  a  slight  slippage 
between  the  bottom  and  top  rolls,  but  in  case  the  top  - 
roll  is  properly  constructed,  weighted,  and  lubricated,  this 
slippage  is  very  slight  and  will  be  ignored  in  the  following 
calculations. 

The  back  roll,  which  is  the  feed-roll,  always  rotates  at  the 
slowest  speed  and  the  front  roll  at  the  highest,  the  speed  of 
the  other  rolls  being  so  arranged  that  c  revolves  a  little 
more  quickly  than  d,  and  b  still  more  quickly  than  c,  but  at  a 
less  speed  than  a.  The  speed  of  these  rolls  is  usually 
indicated  by  the  number  of  revolutions  per  minute,  often 
abbreviated  to  R.  P.  M.  The  direction  of  rotation  of  the 
rolls  is  shown  by  a  small  arrow  within  the  section  of  each. 

Between  d  and  dt  a  ribbon  of  cotton  is  fed  and  is  carried 
forwards,  as  shown,  between  each  pair  of  rolls,  until  it 
emerges  at  the  front.  The  direction  of  movement  of  this 
ribbon  of  fibers  is  shown  by  the  arrows.  The  upper  rolls 
are  weighted  in  such  a  manner  as  to  firmly  grip  the  fibers 
that  pass  below  them,  and  thus  if  the  spaces  between  the 
centers  of  each  pair  of  rolls  are  properly  adjusted  and  the 
relative  speeds  of  the  rolls  accurately  arranged,  the  principle 
of  drawing  the  fibers  past  one  another  by  means  of  a  firm 


15 


DRAFT  CALCULATIONS 


7 


grip  of  their  fore  ends,  the  rear  ends  trailing  behind,  is 
satisfactorily  adopted.  The  same  conditions  continuously 
exist  in  the  machine,  because  as  the  forward  rolls  pass  fibers 
forwards,  the  rear  rolls  are  supplying  new  ones,  and  the 
results  are  thus  comparatively  even  and  regular. 

The  illustration  shows  the  gradual  attenuation  or  reduction 
in  size  of  the  mass  of  cotton,  owing  to  the  increased  speed 
of  each  pair  of  rolls  over  the  preceding  pair.  It  will  be 
seen  that  if  the  surface  speed  of  the  back  roll  is  60  inches 
per  minute  and  that  of  the  front  roll  360  inches,  the  sliver 
emerging  from  the  front  roll  will  be  six  times  as  long  and 
consequently  six  times  as  fine  as  when  entering  the  back  roll. 

The  arrangement  just  described  is  only  one  of  many  found 
in  cotton-yarn-preparation  machinery  and  is  merely  given  as 
an  example.  Draft  could  be  produced  between  only  two 
pair  of  rolls  almost  contiguous;  again,  these  two  rolls, 
known  as  the  feed-roll  and  delivery  roll,  respectively,  might 
have  between  them  a  large  number  of  other  rolls,  or  a  num¬ 
ber  of  cylinders  or  rollers,  or  other  means  of  producing 
draft,  but  the  draft  would  be  computed  between  the  feed-rolls 
and  the  delivery  rolls  if  the  total  draft  were  desired. 

6.  Methods  of  Finding  Draft. — Draft  is  the  ratio  of 
the  speed  of  the  delivery  to  that  of  the  feed  part  of  a 
machine.  It  indicates  the  ratio  between  the  surface  speed  of 
the  front,  or  delivery,  roll  and  the  surface  speed  of  the  back, 
or  feed,  roll,  and  may  be  found  in  different  ways. 

1.  Draft  may  be  found  by  dividing  the  space  moved 
through  in  a  given  time  by  a  point  on  the  surface  of  the 
feed-roll,  into  the  space  moved  through  in  the  same  time  by 
a  point  on  the  surface  of  the  delivery  roll. 

Example  1. — Referring  to  Fig.  1,  if  a  point  on  the  surface  of  d 
moves  3  inches  while  a  point  on  the  circumference  of  the  roll  a  moves 
18  inches,  what  is  the  draft? 

Solution. —  18  -=-  3  =  6,  draft.  Ans. 

2.  Draft  may  be  found  by  dividing  the  weight  per  yard 
of  the  product  delivered,  into  the  weight  per  yard  of  the 
material  fed  into  the  feed-rolls. 


8 


DRAFT  CALCULATIONS 


§15 


Example  2. — If  1  yard  of  cotton  fed  into  the  machine  weighs 
72  grains,  and  1  yard  delivered  at  the  front  of  the  machine  weighs 
12  grains,  what  is  the  draft? 

Solution. —  72  -p  12  =  6,  draft.  Ans. 

3.  Draft  may  be  found  by  dividing  the  number  of  yards 
delivered  by  the  delivery  roll  in  a  certain  time,  by  the  num¬ 
ber  of  yards  fed  into  the  feed-roll  in  the  same  time. 

Example  3. — If  2  yards  of  cotton  is  fed  into  a  machine  during  the 
same  time  that  12  yards  is  delivered,  what  is  the  draft? 

Solution.—  12  -p  2  =  6,  draft.  Ans. 

It  will  be  observed  that  these  three  methods  of  finding  the 
draft  deal  with  the  ratio  between  the  length,  weight,  or  speed 
of  the  material  fed  and  the  corresponding  condition  of  the 
material  delivered;  and  from  these  examples  will  be  deduced 
the  facts  that  while  the  length  of  material  fed  into  the 
machine  is  increased  by  drafting,  the  weight  per  unit  of 
length  is  always  decreased  in  the  same  proportion.  The 
problems  just  presented  are  comparatively  easy  to  under¬ 
stand,  but  the  data  necessary  to  solve  the  problem  of  draft 
in  any  of  the  methods  just  shown  is  not  always  available. 
It  may  be  necessary  to  calculate  the  draft  before  a  machine 
is  put  into  operation  or  without  being  able  to  ascertain  the 
weight  or  length  of  material  fed  or  delivered,  and  it  is  then 
necessary  to  find  the  draft  and  perform  various  other  draft 
calculations  by  a  consideration  of  the  gearing  that  connects 
the  rolls  and  also  the  sizes  of  the  rolls  themselves.  This 
is  the  most  common  method  of  providing  data  for  draft 
calculations. 

Draft  may  therefore  be  defined  in  various  ways,  with  ref¬ 
erence  to  the  machine  or  part  of  a  machine  in  which  draft  is 
being  produced:  (1)  The  ratio  between  the  length  delivered 
and  the  length  fed  in  a  certain  time;  (2)  the  ratio  of  speed 
between  a  point  on  the  delivery  roll  and  a  point  on  the 
feed-roll;  (3)  the  number  of  times  that  a  certain  length  of 
material  is  increased  while  being  operated  on;  (4)  the  ratio 
between  the  weight  of  a  certain  length  of  material  fed  and 
the  weight  of  the  same  length  of  material  delivered;  (5)  the 


§15 


DRAFT  CALCULATIONS 


9 


number  of  times  that  the  weight  of  a  certain  length  of 
material  is  decreased  while  being  operated  on. 

Each  of  these  definitions  has  the  same  meaning,  the  same 
facts  being  stated  in  different  ways.  The  different  methods 
of  finding  draft  are  each  based  on  one  of  these  definitions, 
according  to  the  data  provided  or  available. 


GEARING  OF  ROLES 

7.  Figs.  2  and  3  represent  different  views  of  four  pair  of 
rolls  and  their  gearing,  Fig.  2  showing  the  principal  driving 
end.  The  figures  show  the  front,  second,  third,  and  back 


Fig.  2 

top  rolls,  and  the  front  bottom  roll.  The  front  rolls  are 
marked  a  and  ax;  the  second  top  roll,  d1;  the  third,  c and  the 
back  top  roll,  dx.  The  bottom  roll  a  drives  the  back  bottom 
roll  by  a  train  of  gears  e,  /,  g,  h;  e  is  on  the  roll  a;  h  is  on 
the  back  roll;  /  and  g  are  compounded  and  revolve  on  a  stud. 
The  third  bottom  roll  is  driven  from  the  back  roll  by  means 
of  three  gears  j,  k ,  /,  Fig.  3;  j  is  on  the  back  roll;  l  is  on  the 
third  roll;  while  k  is  an  idler,  or  carrier,  gear  revolving  on  a 
stud.  The  second  bottom  roll  is  driven  from  the  roll  a  by 


10 


DRAFT  CALCULATIONS 


§15 


means  of  three  gears  m,  n ,  o;  m  is  on  the  second  roll;  o  is  on 
the  front  roll  a ;  while  n  is  a  carrier  gear  revolving  on 
a  stud. 

A  carrier  gear  is  usually  placed  between  a  driver  and  a 
driven  gear  when  it  is  not  convenient  to  make  them  large 
enough  to  mesh  with  each  other,  or  where  it  is  necessary  to 
change  the  direction  of  motion  of  the  driven  gear  without 
changing  its  speed.  It  is  important,  in  connection  with 
draft  calculations,  to  notice  which  gears  are  merely  carrier 
gears,  as  a  carrier  gear  does  not  affect  the  speed,  and  can 
be  left  out  of  all  calculations  of  trains  of  gears  of  which  it 
forms  a  unit. 

The  sizes  of  the  rolls  shown  in  Figs.  2  and  3  are  as 


follows:  Front  roll  a.  If  inches;  second  roll,  If  inches;  third 
roll,  li  inches;  fourth  roll,  li  inches.  These  dimensions 
represent  the  diameter  of  the  roll  in  each  case. 

Although  Figs.  2  and  3  show  a  view  of  four  pair  of  draft 
rolls  mounted  on  stands  as  they  would  appear  in  actual 
operation,  it  would  be  unnecessary  and  inconvenient  in  mill 
work  to  make  a  sketch  of  this  kind  in  order  to  illustrate  the 
various  methods  of  combining  and  driving  draft  rolls.  The 
simplest  method  of  showing  draft  rolls  and  their  gearing, 


15 


DRAFT  CALCULATIONS 


11 


and  the  one  usually  adopted,  is  to  make  a  diagram  in  which 
horizontal  lines  are  drawn  to  show  the  lines  of  rolls,  and  short 
lines  drawn  at  right  angles  to  these  to  indicate  the  gears 
connecting  the  rolls.  It  is  perhaps  better  to  show  the  out¬ 
line  of  the  gears,  instead  of  using  a  line. 

Fig.  4  shows  a  diagram  that  would  represent  the  rolls  and 
gearing  shown  in  both  Figs.  2  and  3.  This  indicates  that 
there  are  four  lines  of  rolls  and  that  the  power  is  received  by 
the  tight  and  loose  pulley  shown  on  the  front-roll  shaft. 
It  further  shows  that  motion  is  conveyed  to  the  back  roll 
from  the  front  roll  by  means  of  the  gears  <?,  /,  g ,  h;  that  the 
third  roll  is  driven  from  the  back  roll  by  means  of  the  gears 
/,  k ,  /;  and  that  the  second  roll  is  driven  from  the  front 


m  54- 


t? 


a 


//' 


h70 


,  fjtt 

/J 


Draff  Change  6ear-g65  — 
/$' 


-* 


eZ2 


Fig.  4 


roll  by  the  gears  n ,  o.  The  number  of  teeth  in  each 

gear  is  shown  in  the  figure,  as  well  as  the  diameters  of 
the  rolls.  The  arrows  indicate  the  places  where  the 
driving  gears  connect  with  the  driven  gears  and  point 
from  the  driving  toward  the  driven  gears. 


DRIVING  AND  DRIVEN  GEARS 

8.  It  is  a  matter  of  great  convenience  in  dealing  with 
calculations  of  drafts  to  be  able  to  refer  to  certain  gears  as 
di’iven  gears  and  others  as  driving  gears,  but  it  is 
frequently  difficult  to  determine  which  are  driven  gears  and 
which  are  driving  gears;  for  trains  of  gears  driving  draft 


12 


DRAFT  CALCULATIONS 


§15 


rolls  are  often  complicated,  as  one  gear  may  transmit  motion 
to  two  trains  of  gears  and  these  in  turn  drive  back  to  other 
trains  of  gears.  In  all  cases  in  connection  with  draft  calcu¬ 
lations,  therefore,  it  is  advisable  to  consider  that  the  gear  on 
the  end  of  the  delivery  roll,  which  transmits  motion  to  the 
other  roll  or  rolls,  is  a  driver,  whether  it  is,  or  is  not,  in 
fact;  and  starting  from  this  point,  the  next  gear  would  there¬ 
fore  be  a  driven,  the  third  a  driver,  the  fourth  a  driven, 
ignoring  carrier,  or  idler,  gears. 

For  example,  if  it  is  desired  to  find  the  draft  between  the 
third  and  back  rolls  in  Fig.  4,  as  only  these  two  rolls  are  to 
be  considered,  the  third  roll  would  be  considered  the  delivery 
roll  and  the  gear  on  it,  viz.,  /,  the  driver,  while  the  gear  on 
the  back  roll,  viz.,/,  must  be  the  driven,  k  being  a  carrier  and 
consequently  left  out  of  the  calculation.  The  fourth  roll 
would  be  considered  the  feed-roll. 


CALCULATING  DRAFT  FOR  COMMON  ROLLS 


9.  Effect  of  the  Diameters  of  Rolls  on  Draft. 
Although  in  reality  the  draft  between  two  pair  of  rolls  rep¬ 
resents  the  ratio  of  the  circumferential  speed  of  one  pair  to 
the  circumferential  speed  of  the  other,  still  it  is  not  neces¬ 
sary  to  take  into  consideration  the  circumference  of  the  rolls 
when  calculating  draft.  For  example,  the  roll  a,  in  Fig.  4, 
has  a  diameter  of  If  inches,  giving  a  circumference  of 
4.3197  ( If  X  3.1416  =  4.3197) ;  the  diameter  of  the  roll  d  is 
li  inches,  giving  a  circumference  of  3.5343  (lix  3.1416  = 
3.5343).  Suppose  that  both  rolls  make  one  revolution  in 
the  same  time;  then  a  point  on  a  would  move  through  4.3197 
inches  while  a  point  on  the  back  roll  was  moving  through 
3.5343  inches;  or,  in  other  words,  4.3197  inches  of  cotton 
would  be  delivered  by  the  roll  a  in  the  same  time  that 
3.5343  inches  was  being  fed  to  the  back  roll  d.  The  draft, 
then,  between  these  two  rolls  would  be  expressed  by  the 
ratio  between  4.3197  and  3.5343,  which  is  equivalent  to  a 


draft  of 


4,3197 

3.5343’ 


1.222. 


However,  the  ratio  between  the 


§15 


DRAFT  CALCULATIONS 


13 


diameters  of  the  rolls,  or  If  and  li,  will  be  found  to  be 
the  same  as  the  ratio  between  the  circumferences  (If  -f-  li 
=  1.22);  consequently,  in  all  draft  calculations  the  diameters 
of  the  rolls  are  considered  and  not  the  circumferences. 

The  sizes  of  rolls  are  usually  expressed  by  their  diameters 
when  they  are  used  in  connection  with  machinery.  It  is 
easier  to  measure  the  diameter  of  a  roll  than  it  is  to  measure 
the  circumference.  If  the  size  of  a  roll  is  said  to  be  li  inches, 
it  is  generally  understood  that  the  roll  is  li  inches  in  diam¬ 
eter.  In  draft  calculations  only  the  sizes  of  the  bottom  rolls 
are  taken  into  account.  The  top  rolls  are  driven  by  frictional 
contact  with  the  bottom  rolls,  and  therefore  should  always 
revolve  at  the  same  circumferential  speed;  consequently,  the 
sizes  of  the  top  rolls  can  be  ignored. 

Another  point  to  be  taken  into  consideration  in  this  con¬ 
nection  is  that  the  diameters  of  draft  rolls  in  cotton  machinery 
are  always  expressed  in  inches  and  fractions  of  an  inch.  It 
is  far  simpler,  when  performing  draft  calculations,  to  change 
the  numbers  representing  the  diameters  of  the  rolls  to  frac¬ 
tions  having  a  common  denominator,  and  then  omit  these 
common  denominators  from  the  calculations.  For  example, 
the  front  roll  in  Fig.  4  would  be  considered  as  a  roll  V-  inches 
in  diameter  (lf  =  J8L),  while  the  back  roll  would  be 
considered  as  a  roll  f  inches  in  diameter  (li  =  -f);  conse¬ 
quently,  if  these  two  rolls  are  making  the  same  number  of 
revolutions  per  minute,  the  draft  between  them  would  be 
represented  by  the  ratio  of  11  to  9,  which  is  equivalent  to  a 
draft  of  -<r,  or  1.222. 

10.  Effect  of  tlie  Size  of  Gears  on  Draft. — The  pre¬ 
vious  example  of  a  draft  calculation  referred  to  a  case  where 
the  draft  was  produced  by  the  difference  in  the  size  of  the 
rolls,  it  having  been  considered  that  the  number  of  teeth  in 
the  gears  was  such  as  to  cause  the  revolutions  per  minute  of 
both  rolls  to  be  equal.  A  calculation  of  draft  will  now  be 
considered  where  the  sizes  of  the  gears  connecting  the  rolls 
affects  the  amount  of  draft,  it  being  assumed  in  this  case  that 
the  rolls  are  of  the  same  diameter.  Taking  the  arrangement 


14 


DRAFT  CALCULATIONS 


§15 


shown  in  Figs.  2  and  3  and  referring  to  the  diagram  of  such 
arrangement  in  Fig.  4,  assuming  that  the  roll  a  is  li  inches 
in  diameter,  and  that  the  roll  d  is  also  li  inches  in  diameter, 
but  that  the  gear  e  contains  22  teeth,  the  gear  /,  98  teeth,  the 
gear  g,  65  teeth,  and  the  gear  h,  70  teeth,  then  the  draft 
between  these  two  rolls  would  be  expressed  by  the  ratio 
between  the  products  of  70  X  98  and  22  X  65,  which  is  equiva¬ 


lent  to  a  draft  of 


70X98 


or  4.797.  Since  the  diameters  of 


22  X  65’ 

the  rolls  are  alike,  this  means  that  while  1  inch  of  cotton 
is  being  fed  to  the  roll  d ,  4.797  inches  is  being  delivered 
by  the  roll  a. 


11.  Combined  Effect  of  Gears  and  Rolls  on  Draft. 
Various  examples  have  now  been  given  of  the  simplest 
forms  of  draft  calculations,  eliminating  complicated  arrange¬ 
ments  as  much  as  possible;  but  in  practice  when  calculating 
drafts  by  means  of  gears,  it  is  found  that  both  the  diameters 
of  the  rolls  and  the  sizes  of  the  gears  must  be  considered, 
and  a  rule  will  now  be  given  that  will  be  found  to  meet 
almost  every  possible  combination  of  gears  and  rolls  of 
which  the  draft  is  required  to  be  calculated. 

To  find  the  draft  between  two  pair  of  rolls  connected  by 
a  train  of  gears: 


Rule. — Always  assume  that  the  gear  on  the  delivery  roll  is  a 
driver;  multiply  all  driven  gears  by  the  diameter  of  the  delivery 
roll ,  expressed  hi  eighths  of  an  inch ,  and  divide  by  the  product 
of  all  the  driving  gears  and  the  diameter  of  the  feed-roll , 
expressed  in  eighths  of  an  inch. 


The  reasons  for  the  formulation  of  this  rule  and  its 
expression  in  the  manner  given  can  be  found  by  a  careful 
study  of  the  arrangement  of  draft  rolls  and  gears  represented 
by  the  diagram,  Fig.  4.  The  application  of  the  rule  to  find¬ 
ing  the  draft  between  a  and  d  would  result  in  the  diameter 
of  the  roll  a  and  the  number  of  teeth  in  the  gears  /  and  h 
being  placed  as  the  numerator  of  a  fraction,  and  the  diameter 
of  the  roll  d  and  the  number  of  teeth  in  the  gears  e  and  g 


15 


DRAFT  CALCULATIONS 


15 


as  the  denominator  of  a  fraction;  consequently,  an  increase 
in  the  diameter  of  the  front  roll  would  cause  the  calculation 
to  show  an  increased  draft.  An  increase  in  the  size  of  the 
gears  /  or  h  would  also  cause  the  calculation  to  show  an 
increased  draft,  while  an  increase  in  the  size  of  the  feed-roll, 
or  an  increase  in  the  size  of  the  gears  e  or  g,  would  cause  the 
calculation  to  show  a  decreased  draft. 

A  careful  study  of  the  figure  proves  that  these  changes 
would  give  in  practice  the  results  indicated.  For  instance, 
assuming  that  the  speed  of  the  front  roll  remains  the  same 
and  its  diameter  is  increased,  the  draft  would  be  increased, 
as  it  would  deliver  more  material  in  the  same  space  of  time. 
An  increase  in  the  size  of  the  back  roll  would  reduce  the 
draft,  because,  all  other  conditions  remaining  the  same,  a 
greater  length  of  material  would  be  fed  to  the  rolls  while  the 
the  same  length  was  being  delivered  at  the  front,  and  conse¬ 
quently  the  draft  operating  on  the  material  must  be  smaller. 
Similarly,  an  increase  in  the  size  of  the  gears  e  or  g  would 
result  in  the  feed-roll  taking  in  more  material  in  the  same 
space  of  time,  consequently  reducing  the  draft;  while  an 
increase  in  the  size  of  the  gears  /  or  h  would  result  in  the 
feed-roll  taking  in  less  material  in  the  same  space  of  time 
and,  as  the  length  delivered  at  the  front  would  remain  the 
same,  the  draft  would  consequently  be  increased. 

12.  The  drafts  between  these  rolls  will  now  be  con¬ 
sidered,  and  the  following  statements,  which  have  been 
previously  made,  should  be  borne  in  mind.  In  finding 
drafts,  the  rule  given  in  Art.  11  should  be  adopted  in  all 
cases  where  it  applies,  as  it  is  the  most  useful  and  reliable 
rule  that  can  be  given  for  draft  calculations.  The  gear  on 
the  delivery  roll  should  always  be  considered  as  a  driver, 
and  counting  from  this  gear  the  next  gear  will  be  a  driven, 
and  so  on  alternately  throughout  the  train  of  gears,  always 
providing  that  the  carrier  gears  in  the  train,  if  any,  are 
ignored  in  consequence  of  their  being  simply  idlers  and  not 
affecting  the  amount  of  draft,  no  matter  how  many  or 
how  few  teeth  they  contain.  The  delivery  roll  should  be 


16 


DRAFT  CALCULATIONS 


§15 


understood  as  the  front  roll  of  those  rolls  that  are  under 
consideration.  If  the  draft  is  being  figured  between  a 
and  d,  Fig.  4,  a  is  the  delivery  roll;  if  between  b  and  c,  b  is 
considered  the  delivery  roll. 

In  the  combination  of  rolls  shown  in  Fig.  4,  it  is  possible 
to  calculate  four  different  drafts:  a ,  the  total  draft,  which 
represents  the  extent  of  attenuation  between  the  back  roll 
and  the  front  roll;  b,  the  draft  between  the  front  roll  and  the 
second;  c,  the  draft  between  the  second  and  third  rolls;  and 
d,  the  draft  between  the  third  and  fourth  rolls.  The  accu¬ 
racy  of  the  calculation  for  the  total  draft  a  can  always  be 
proved  by  multiplying  the  other  three  drafts  b,  c,  d 
together,  although  owing  to  the  fact  that  the  separate  calcu¬ 
lations.  may  not  come  out  even  in  three  decimal  places,  the 
draft  obtained  in  the  proof  may  not  exactly  correspond  with 
the  total  draft  as  obtained  at  first.  In  this  connection  it  is 
advisable  to  emphasize  the  fact  that  the  intermediate  drafts 
should  be  multiplied  and  not  added.  This  is  an  error  fre¬ 
quently  made  by  beginners,  as  they  assume  that  by  adding 
together  the  drafts  between  the  first  and  second,  the  second 
and  third,  and  the  third  and  fourth  rolls,  the  total  draft  is 
thus  obtained;  but  this  is  not  so,  as  these  intermediate 
drafts  must  be  multiplied  together  in  order  to  find  the  draft 
that  agrees  with  the  total  draft  from  front  to  back. 

Example  1. — Referring  to  Fig.  4,  the  front  roll  is  If  inches  in 
diameter  and  carries  a  22-tooth  gear  driving  a  98-tooth  gear.  Com¬ 
pounded  with  this  is  a  65  gear  driving  a  70-tooth  gear  on  the  back  roll, 
which  is  If  inches  in  diameter.  What  is  the  total  draft,  or  the  draft 
between  the  front  and  back  pairs  of  rolls? 

Solution. —  =  5.863,  total  draft.  Ans. 

9  X  22  X  65 

Example  2. — Referring  to  Fig.  4,  the  front  roll  is  If  inches  in 
diameter  and  carries  an  18-tooth  gear  driving  a  54  on  the  second  roll, 
which  is  also  If  inches  in  diameter.  What  is  the  draft  between  these 
two  pair  of  rolls? 

0  11  X  54  _  _  .  . 

Solution. —  —  =  3,  draft.  Ans. 

11  x\  lo 

Example  3. — Referring  to  Fig.  4,  the  second  roll  is  If  inches  in 
diameter  and  carries  a  54-tooth  gear  driving  an  18  on  the  front  roll. 


§15 


DRAFT  CALCULATIONS 


17 


On  the  other  end  of  the  front  roll  is  a  22  driving  a  98  compounded 
with  a  65,  which  drives  a  70  on  the  back  roll.  On  the  other  end  of  the 
back  roll  is  a  40  driving  a  30  on  the  third  roll,  which  is  li  inches  in 
diameter.  What  is  the  draft  between  the  second  and  the  third  rolls? 


Solution. — 


11  X  18  X  98  X  70  X  30 
9  X  54  X  22  X  65  X  40 


1.466,  draft.  Ans. 


Example  4. — The  third  roll  in  Fig.  4  is  driven  from  the  back  roll. 
The  back  roll  is  li  inches  in  diameter  and  carries  a  40-tooth  gear  dri¬ 
ving  a  30  on  the  third  roll,  which  is  also  li  inches  in  diameter.  What 
is  the  draft  between  these  two  pair  of  rolls? 

q  v  40 

Solution. —  ■  =  1.333,  draft.  Ans. 

y  /\  ou 


Proof. — The  total  draft  as  found  in  example  1  may  be 
proved,  as  already  stated,  by  multiplying  together  the  drafts 
obtained  in  examples  2,  3,  and  4. 

3  X  1.466  X  1.333  =  5.862 

Owing  to  the  slight  slippage  previously  mentioned  between 
the  bottom  and  top  rolls  of  each  pair,  no  rule  for  finding  the 
draft  would  give  absolutely  accurate  results  when  applied  to 
common  rolls.  The  slippage  is  usually  so  slight  and  the 
effect  of  the  slippage  in  one  pair  of  rolls  is  so  nearly  neutral¬ 
ized  by  the  effect  of  the  slippage  in  each  of  the  other  pai/s 
of  rolls,  that  the  resultant  slippage  is  generally  infinitesimal 
and  can  be  ignored.  The  rules  that  are  given  are  as  nearly 
accurate  as  is  required  for  all  practical  purposes. 


BREAK  DRAFT 

13.  An  expression  frequently  found  in  cotton  mills  in 
connection  with  drafts  is  that  of  break  draft.  While  the  con¬ 
sideration  of  the  break  draft  is  not  of  much  practical  impor¬ 
tance,  yet  a  brief  description  of  the  subject  is  advisable. 
Break  draft  is  a  draft  between  two  contiguous  pairs  of  rolls 
that  are  not  directly  connected  by  means  of  gears.  Refer¬ 
ence  to  Fig.  4  shows  that  the  second  and  third  pair  of  rolls 
are  adjacent  to  each  other,  and  yet  not  directly  connected  by 
means  of  gears,  the  driving  of  the  third  being  attained  by 


18 


DRAFT  CALCULATIONS 


15 


means  of  a  long  train  of  gears  from  the  delivery  roll,  while 
the  second  roll  is  driven  by  a  short  train  of  gears  from  the 
delivery  roll.  The  break  draft  in  this  case,  therefore,  is 
found  between  the  second  and  third  pair  of  rolls. 

Break  draft  may  be  found  in  two  ways,  one  method  being 
to  start  with  the  gear  m ,  Fig.  4,  and  end  with  the  gear  /, 
using  the  diameters  of  the  rolls  b  and  c. 

The  second  method  is  to  calculate  the  draft  between  the  first 
and  fourth  rolls,  Fig.  4;  then  between  the  third  and  fourth; 
and  next  between  the  first  and  second  rolls.  The  drafts 
between  the  third  and  fourth  and  the  first  and  second  rolls 
are  multiplied  together  and  divided  into  the  draft  between 
the  first  and  fourth  rolls,  or  the  total  draft.  The  quotient 
will  be  the  break  draft,  or  the  draft  between  the  second  and 
third  rolls. 

Example. — What  is  the  break  draft,  or  the  draft  between  the  second 
and  third  rolls  shown  in  Fig.  4? 

Solution  (a). — Figured  according  to  the  first  method, 


11  X  18  X  98  X  70  X  30 
9  X  54  X  22  X  65  X  40 


=  1.466,  break  draft. 


Ans. 


Solution  (5). — Figured  according  to  the  second  method, 


9X40 

9X30 


=  1.333,  draft  between  third  and  fourth  rolls 


11  X54 
11  X  18 


3,  draft  between  first  and  second  rolls 


11  X  98  X  70 
9  X  22  X  65 


5.863,  total  draft, 


1.333  X3  =  3.999;  5.863  -p  3.999  =  1.466,  break  draft.  Ans. 


14.  Fig.  5  shows  four  pair  of  drawing  rolls  geared  in  a 
different  manner  from  that  shown  in  Fig.  4.  The  front  roll 
a  receives  motion  through  the  tight  pulley.  The  gear  e  on 
this  roll  drives  the  third  roll  c  by  means  of  the  gears  /,  g,  h\ 
the  fourth  roll  d  is  driven  from  the  third  roll  by  the  gears 
/,  k ,  /;  k  is  an  idler,  or  carrier,  gear.  The  second  roll  b  is 
driven  from  the  third  roll  by  the  gears  /,  m ,  n;  the  gear  m 
is  an  idler,  or  carrier,  gear.  The  break  draft  in  this  case  is 
located  between  the  first  and  second  rolls. 


15 


DRAFT  CALCULATIONS 


19 


Fig.  6  shows  another  method  of  gearing  a  set  of  drawing 
rolls.  The  front  roll  a  receives  motion  through  the  tight 
pulley,  and  the  gear  e  on  this  roll  drives  the  back  roll  d 


Draft  Change  Gear—g8/- 

.3" 

t£ 

e20 

Fig.  5 


through  the  gears  /,  g,  h.  The  gear  j  on  the  back  roll  drives 
the  third  roll  c  through  the  gears  k ,  /;  k  is  a  carrier  gear. 


a 


/b 


D60 


/r 


/r 


n  20 

Draft  Change  Gear-g  78 


/  J  f- 

/s 


f//5 


Fig.  6 


The  second  roll  b  is  driven  from  the  third  roll  by  means  of 
the  gears  /,  m,  n;  m  is  a  carrier  gear.  The  break  draft  in 
this  figure  is  located  between  the  first  and  second  rolls. 


20 


DRAFT  CALCULATIONS 


§15 


METALLIC  ROLLS 


15.  In  recent  years  what  are  known  as  metallic  rolls 
have  been  introduced,  especially  on  the  preparatory 
machines  in  the  processes  of  cotton-yarn  preparation,  such 


as  railway  heads,  drawing  frames,  and  slubbers.  Owing  to 
the  peculiarity  of  construction  of  these  rolls,  the  rules 


previously  given  for  figuring  draft  do  not  apply  without 
modification.  They  have  both  the  upper  and  lower  rolls 
constructed  of  steel,  and  both  rolls  are  fluted  longitudinally. 


15 


DRAFT  CALCULATIONS 


21 


These  flutes  are  different  in  shape  and  considerably  coarser 
than  the  flutes  in  common  steel  rolls,  and  when  in  operation 
the  flutes  of  one  roll  project  into  the  flutes  of  the  other  roll, 
being  prevented  from  coming  into  too  close  contact  by 
means  of  collars  on  the  rolls. 

Fig.  7  shows  a  general  view  of  a  set  of  metallic  rolls  in 
position,  Fig.  8  giving  a  view  of  the  ends  of  two  rolls; 
b  and  bt  are  the  fluted  portions  of  the  bottom  and  top  rolls, 
respectively,  meshing  into  one  another;  a  and  are  the 
collars  on  the  rolls,  which  prevent  the  flutes  from  bottoming. 


The  collars  are  slightly  smaller  than  the  outside  diameter  of 
the  boss,  which  is  the  name  applied  to  each  fluted  portion  of 
the  rolls,  and  thus  provides  for  a  certain  amount  of  inter¬ 
locking  of  the  bosses.  A  section  through  a  portion  of  the 
two  rolls  is  shown  in  Fig.  9,  the  same  letters  in  both  Figs.  8 
and  9  applying  to  the  same  parts.  The  sliver  c  as  operated 
on  by  the  rolls  is  also  indicated. 

As  the  fluted  portions  of  the  rolls  are  not  actually  in 
contact,  the  individual  fibers  in  the  sliver  are  free  to  slide 


22 


DRAFT  CALCULATIONS 


§15 


past  one  another,  and  as  the  tooth  e,  Fig.  9,  falls  into  the 
space  d  as  the  roll  revolves,  gradually  assuming  the  same 
position  as  e1  within  the  space  du  an  attenuation  is  produced 
in  the  sliver,  lengthening  it  considerably.  The  exact  per¬ 
centage  of  this  increase  of  length  or  drafting  varies  accord¬ 
ing  to  the  fineness  or  coarseness  of  the  flutes  and  the  weight 
per  yard  of  the  sliver,  coarse  flutes  producing  a  greater  draft, 
and  a  lighter  sliver  being  more  susceptible  to  the  influence 
of  the  drafting  action  of  the  teeth  of  the  rolls.  This  draft¬ 
ing  action  in  each  pair  of  rolls  is  irrespective  of  the  drafting 
effect  of  the  relative  speed  of  each  two  pair  of  rolls  in 
the  set,  and  allowance  must  be  made  for  it  in  making  draft 
calculations,  a  certain  percentage  being  added  to  the  calcu¬ 
lated  draft  in  order  to  find  the  actual  draft. 

16.  Allowances  Made  In  Calculating  Production 
and  Draft. — Another  feature  of  metallic  rolls  is  that  in 
calculating  the  amount  of  material  fed  into  a  machine  or 
delivered  by  a  machine  equipped  with  these  rolls,  the  actual 
outside  diameter  multiplied  by  3.1416  will  not  give  the 
amount  of  material  fed  into  or  delivered  by  a  roll  in  one 
revolution.  The  crimping  action  of  the  rolls  causes  a  greater 
length  to  be  fed  and  delivered  than  would  be  the  case  with 
common  rolls  of  the  same  diameter.  It  is  usually  assumed 
that  one-third  more  material  is  delivered  by  a  metallic  roll 
than  by  a  common  roll  of  the  same  diameter  on  this  account, 
the  zigzag  lines  of  the  circumference  being  about  33-3  per 
cent,  longer  than  the  circumference  of  a  circle  passing 
through  the  points  of  the  teeth.  To  obtain  accurate  results 
in  figuring  production  with  metallic  rolls,  therefore,  a  certain 
percentage  — usually  333  —  must  be  added  to  the  diameter  of 
each  roll.  A  1-inch  roll  would  be  taken  as  1.33  inches; 
li-inch,  1.5  inches;  li-inch,  1.67  inches;  ll-inch,  1.83  inches; 
li-inch,  2  inches.  The  foregoing  allowances  are  for  ordinary 
metallic  rolls  constructed  with  32  flutes  for  each  inch  of 
diameter.  Metallic  drawing  rolls  are  made  with  flutes  of 
varying  pitch,  either  16  pitch,  24  pitch,  or  32  pitch.  This 
means  that  for  each  inch  of  diameter  of  the  roll  there  are 


15  • 


DRAFT  CALCULATIONS 


23 


either  16,  24,  or  32  flutes.  For  instance,  a  li-inch  roll  of  32 
pitch  would  have  40  flutes  in  its  circumference.  The  allow¬ 
ance  of  33i  per  cent,  is  made  in  case  of  rolls  being-  con¬ 
structed  of  32  pitch,  but  for  rolls  of  16  pitch  this  allowance 
is  increased  to  50  per  cent.,  and  for  24  flutes  to  the  inch,  an 
allowance  of  40  per  cent,  is  made. 

Another  feature  to  consider  in  connection  with  metallic 
rolls  is  that  the  extent  of  the  crimping  action  or  attenuation 
through  the  interlocking  of  the  rolls  is  less  for  heavy  slivers 
than  for  light  slivers,  as  heavy  slivers  resist  the  tendency  of 
the  rolls  to  interlock,  and,  in  some  cases  where  they  are 
insufficiently  weighted,  will  raise  the  top  roll  and  pass 
through  in  almost  a  straight  line.  It  therefore  follows  that 
the  drafting  action  is  greater  with  light  slivers  than  with 
heavy  ones,  and  that  if  the  front  and  back  rolls  of  the 
machine  are  both  the  same  pitch  in  the  flutes,  the  drafting 
action  of  the  back  pair  of  rolls  is  less  than  that  of  the  front 
pair,  since  the  sliver  becomes  thinner  as  it  passes  forwards 
through  the  machine,  on  account  of  being  acted  on  by  the 
draft  between  each  successive  pair  of  rolls. 

17.  The  action  of  metallic  rolls  as  compared  with  common 
rolls  may  be  described  as  follows,  assuming  that  a  compar¬ 
ison  is  being  made  between  a  set  of  four  pair  of  common 
and  four  pair  of  metallic  rolls  all  of  the  same  outside  diam¬ 
eter,  all  geared  in  the  same  manner,  and  all  running  at  the 
same  speed.  The  back  metallic  rolls  would  absorb  approx¬ 
imately  25  per  cent,  more  material  fed  into  them  and  deliver 
approximately  33i  per  cent,  more  material  than  the  common 
rolls.  This  is  due  to  the  fact  that  the  crimping  action  of 
the  back  rolls  acting  on  a  heavy  sliver  just  as  it  is  fed  into 
the  machine  does  not  have  so  great  an  effect  as  the 
crimping  action  of  the  front  rolls,  which  are  acting  on  a  fine 
sliver  that  has  been  drawn  out  to  probably  six  times  its 
original  length  on  account  of  being  acted  on  by  the  draft 
between  the  rolls.  In  this  case,  therefore,  the  draft  of  the 
metallic  rolls  would  have  to  be  figured  in  the  ordinary  way, 
as  for  common  rolls,  and  an  addition  of  33-3  per  cent,  minus 


24 


DRAFT  CALCULATIONS 


§15 


25  per  cent.,  equaling  8}  per  cent.,  made  to  the  calculated 
draft  so  as  to  equal  the  actual  draft  in  the  case  of  the  metallic 
rolls. 

In  cases  where  the  sliver  is  between  45  and  70  grains  in 
weight,  the  draft  between  4i  and  7,  the  back  and  front  rolls 
approximately  of  the  same  size,  and  flutes  with  a  32  pitch 
used,  an  allowance  of  9  per  cent,  over  and  above  the  draft  as 
calculated  with  common  rolls  is  frequently  made,  in  order  to 
arrive  at  the  actual  draft  in  case  of  metallic  rolls. 

From  the  preceding  statements  it  will  be  seen  that  this 
allowance  cannot  be  arbitrary.  The  allowance  should  be 


TABLE  I 


Weight  of  Sliver 

Light  Draft 

Medium  Draft 

Heavy  Draft 

50-grain  sliver 

8  per  cent. 

10  per  cent. 

12 

per  cent. 

60-grain  sliver 

7  per  cent. 

9  per  cent. 

1 1 

per  cent. 

70-grain  sliver 

6  per  cent. 

8  per  cent. 

10 

per  cent. 

80-grain  sliver 

5  per  cent. 

7  per  cent. 

9 

per  cent. 

90-grain  sliver 

4  per  cent. 

6  per  cent. 

8 

per  cent. 

100-grain  sliver 

32  per  cent. 

5-2  per  cent. 

7 

per  cent. 

1 10-grain  sliver 

3  per  cent. 

5  per  cent. 

7 

per  cent. 

120-grain  sliver 

3  per  cent. 

4-2  per  cent. 

6 

per  cent. 

130-grain  sliver 

22  per  cent. 

4  per  cent. 

52 

per  cent. 

140-grain  sliver 

2I  per  cent. 

32  per  cent. 

5 

per  cent. 

150-grain  sliver 

2  per  cent. 

3  per  cent. 

4 

per  cent. 

increased  in  case  of  running  very  light  slivers,  in  case  of 
rolls  being  used  of  coarser  pitch  than  32,  in  case  of  there 
being  a  heavy  draft  in  the  machines,  or  where  the  front  rolls 
are  very  much  larger  than  the  back  rolls.  The  allowance  is 
materially  reduced  in  case  of  a  heavy^  sliver  being  run 
through  the  machine,  in  case  of  a  light  calculated  draft,  or 
in  case  of  the  back  rolls  being  larger  than  the  front  rolls. 

The  numerous  causes  of  variation  of  the  allowances  render 
it  almost  impossible  to  accurately  figure  drafts  for  metallic 
rolls,  and  in  making  changes  in  machines  fitted  with  metallic 


15 


DRAFT  CALCULATIONS 


25 


rolls  or  in  starting  up  such  machines,  it  is  necessary  to 
experiment  somewhat  with  different  gears  to  arrive  at  the 
desired  result;  but  when  this  result  is  once  obtained,  and  so 
long  as  the  conditions  remain  the  same,  the  results  from 
metallic  rolls  are  just  as  regular  as  from  common  rolls. 
Table  I  gives  the  allowances  that  should  be  made,  under 
various  conditions,  on  the  calculated  draft  for  common  rolls 
in  order  to  ascertain  what  the  draft  would  be  if  metallic  rolls 
of  the  same  diameter  were  used  and  assuming  that  the  front 
and  back  rolls  do  not  vary  greatly  in  diameter. 

The  table  must  not  be  taken  as  arbitrary,  for  slight  varia¬ 
tions  from  this  must  be  expected  in  practice.  Drafts  from 
5  to  7  may  be  considered  medium  drafts. 


Example. — Referring  to  Fig.  4,  what  is  the  total  draft  both  for 
common  and  for  metallic  rolls,  assuming  that  a  60-grain  sliver  is 
passing  into  the  machine? 


Solution. — 


11  X  98  X  70 


9  X  22  X  65 
9%  of  5.863  =  .52767. 

5.863  +  .52767  =  6.39067,  draft  for  metallic  rolls 


=  5.863,  draft  for  common  rolls.  Ans. 


Ans. 


DRAFTING  WITH  SPECIAL  REFERENCE  TO 

THE  MILL 


DRAFT  GEARS 

18.  In  each  principal  train  of  gears  connecting  draft  rolls, 
one  gear  is  always  spoken  of  as  the  change  gear  or  draft 
gear,  and  this  is  the  one  that  is  usually  changed  under  all 
ordinary  circumstances  for  producing  different  sizes  of 
product  in  the  machine,  or  for  maintaining  the  product  at  the 
same  size  in  case  of  different  weights  of  material  having  to 
be  fed  in  the  back  of  the  machine.  The  draft  gear  is  shown 
in  Figs.  2  and  4,  being  marked  g\  it  is  also  indicated  in 
Figs.  5  and  6. 

The  draft  gear  for  a  set  of  rolls  is  usually  situated  on  a 
stud  together  with  another  gear,  which  is  known  as  the 
crown  gear  in  order  to  distinguish  it  from  the  draft  gear. 


26 


DRAFT  CALCULATIONS 


§15 


Thus,  in  Figs.  4,  5,  and  6  the  draft  gears  g  are  in  every 
instance  placed  on  a  stud  with  the  crown  gears  /.  Crown 
gears  are,  as  a  rule,  the  larger  of  the  two  gears  on  the  stud 
and  stand  higher  than  the  other  gears. 

Any  change  in  the  draft  gear  alters  the  speed  of  the  feed- 
rolls,  while  the  speed  of  the  front  rolls  remains  constant. 
Usually,  a  larger  draft  gear  will  increase  the  speed  of  the 
back,  or  feed,  rolls,  thus  producing  less  draft,  because  more 
cotton  is  being  fed  while  there  has  been  no  change  in  the 
length  of  the  amount  delivered.  Usually,  a  smaller  draft 
gear  will  produce  more  draft,  because  there  will  be  less 
cotton  fed  in,  on  account  of  the  speed  of  the  feed-rolls  being 
reduced,  while  the  same  length  of  cotton  is  being  delivered 
all  the  time.  Thus,  if  it  is  desired  to  produce  more  draft,  the 
speed  of  the  feed-rolls  is  reduced,  while,  on  the  other  hand,  if 
it  is  desired  to  reduce  the  draft,  the  speed  of  the  feed-rolls 
is  increased.  In  almost  every  case  where  draft  is  introduced 
in  cotton-mill  machinery,  this  method  will  hold  good. 

It  should  also  be  noted  that  a  change  in  the  draft  gear^, 
Fig.  4,  makes  no  difference  in  the  ratio  of  speed  between  the 
first  and  second  rolls  or  between  the  third  and  fourth  rolls, 
but  it  does  between  the  first  and  fourth  and  between  the  sec¬ 
ond  and  third  rolls.  This  is  also  true  in  regard  to  Figs.  5 
and  6;  that  is,  any  change  in  the  draft  change  gear  g  will 
only  change  the  draft  between  the  sets  of  rolls  where  the 
break  draft  is  located  and  between  the  first  and  fourth  rolls, 
or  between  the  front  and  back  rolls. 

19.  The  following  rules  apply  to  drafts  and  draft  gears 
when  the  draft  gear  is  a  driver,  assuming  that  the  gear  on 
the  front  roll  is  a  driver. 

To  find  the  draft  gear  required  to  give  a  certain  draft  when  the 
draft  gear  being  used  and  the  draft  being  produced  are  known: 

Rule  I. — Multiply  the  draft  gear  being  used  by  the  draft 
being  produced  and  divide  the  product  by  the  draft  desired. 

Example  1. — Referring  to  Fig.  4,  a  draft  gear  of  65  teeth  produces 
a  draft  of  5.86.  What  draft  gear  will  be  required  to  produce  a  draft 
of  7? 


15 


DRAFT  CALCULATIONS 


27 


Solution. — 


65  X  5.86 
7 


54.41,  a  54  or  55  draft  gear.  Ans. 


To  find  what  draft  a  certain  draft  gear  will  produce  when 
the  draft  gear  being  used  and  the  draft  being  produced  are 
known: 


Rale  II. — Multiply  the  draft  gear  being  used  by  the  draft 
being  produced  and  divide  the  product  by  the  draft  gear  to  be 
used. 

Example  2. — Referring  to  Fig.  4,  a  draft  gear  of  65  teeth  produces 
a  draft  of  5.86.  What  draft  will  a  54  draft  gear  produce? 

Solution. —  65X5^ 86  _  7  draft.  Ans. 

Ot 

20.  The  following  rules  apply  to  drafts  and  draft  gears 
when  the  draft  gear  is  a  driven,  and  for  the  purpose  of  illus¬ 
tration  the  gear  /,  Fig.  4,  which  is  a  driven  gear,  will  be 
considered  as  the  draft  change  gear. 

To  find  the  draft  gear  required  to  give  a  certain  draft  when 
the  draft  gear  being  used  and  the  draft  being  produced  are 
known: 

Rale  I. — Multiply  the  draft  gear  being  used  by  the  draft  to 
be  produced  a?id  divide  the  product  thus  obtained  by  the  draft 
being  produced. 

Example  1. — Referring  to  Fig.  4,  a  draft  of  5.86  is  being  produced 
with  a  98-tooth  draft  gear.  What  draft  gear  will  be  required  to  give  a 
draft  of  7? 

98  v  7 

Solution. —  ■'  - =  117.06  =  117,  draft  gear.  Ans. 

5.8b 


To  find  what  draft  a  certain  gear  will  give  when  the  draft 
gear  being  used  and  the  draft  that  it  is  producing  are  known: 


Rale  II. — Multiply  the  draft  that  is  being  produced  by  the 
draft  gear  that  is  to  be  used  and  divide  the  product  thus  obtained 
by  the  draft  gear  being  used. 


Example  2.— Referring  to  Fig.  4,  a  draft  of  5.86  is  being  produced 
with  a  98  draft  gear.  What  draft  will  be  produced  with  a  117  draft 
gear? 


5.86  X  117 


6.996,  draft.  Ans. 


Solution. — 


98 


28 


DRAFT  CALCULATIONS 


15 


CONSTANTS 

21.  When  a  new  machine  is  started  up  for  the  first  time, 
it  is  necessary  to  calculate  the  draft  and  draft  gears  required 
to  give  a  certain  weight  of  cotton.  These  results  are  obtained 
in  the  manner  previously  described.  All  the  machines  in  a 
card  room  or  spinning  room  have  a  certain  draft  and  these 
drafts  bear  a  certain  relation  to  each  other,  in  order  to  reduce 
the  cotton  to  the  desired  size  and  weight  to  make  yarn. 
After  a  machine  has  been  started,  these  methods  of  finding 
the  draft  and  draft  gear  may  be  shortened  by  employing 
what  are  known  as  constants. 

22.  Constant  Dividends  and  Constant  Factors. 
Constants  are  almost  always  used  in  order  to  shorten  draft 
calculations.  There  are  two  kinds  of  constants  used  in  draft 
calculations;  namely,  constant  dividends  and  constant  factors. 
A  constant  dividend  is  a  number  which,  when  divided  by 
the  draft,  will  give  the  necessary  draft  gear;  or  it  may  be 
defined  as  a  number  which,  when  divided  by  the  draft  gear 
being  used  on  a  machine,  will  give  the  amount  of  draft  that  the 
machine  is  producing.  A  constant  factor  is  a  number  which, 
when  divided  into  the  draft,  will  give  the  draft  gear  necessary 
to  produce  the  desired  draft;  or  it  may  be  defined  as  a  number 
which,  when  multiplied  by  the  draft  gear  being  used  on  a 
machine,  will  give  the  draft  that  the  machine  is  producing. 

Each  different  make  of  machine  and  each  different  kind  of 
machine  has  a  different  constant. 

Assuming  that  the  gear  on  the  front  roll  is  a  driver,  the 
following  statements  may  be  made: 

When  the  draft  gear  is  a  driver,  the  constant  is  always  a 
constant  dividend. 

When  the  draft  gear  is  a  driven,  the  constant  is  always  a 
constant  factor. 

To  find  the  draft  constant  of  a  machine: 

Rule  I. — Perform  the  calculations  exactly  the  same  as  when 
finding  the  draft ,  always  considering  the  draft  gear  as  a 
1-tooth  gear. 


15 


DRAFT  CALCULATIONS 


29 


Example  1. — The  machine  in  Fig.  4  contains  the  following  train  of 
gears:  The  front  roll  a  is  If  inches  in  diameter  and  carries  the 
22-tooth  gear  e  driving  the  98-tooth  gear  f.  Compounded  with  f  is  the 
65  draft  gear  g  driving  the  70-tooth  gear  li  on  the  back  roll  d,  which  is 
1|  inches  in  diameter.  What  is  the  constant  dividend? 

Solution. —  !!  o~r~  =  381,  constant  dividend.  Ans. 

To  find  the  draft  when  the  constant  dividend  and  draft 
gear  are  known: 

Rule  II. — Divide  the  constant  dividend  by  the  draft  gear. 

Example  2. — What  is  the  total  draft  for  Fig.  4  with  a  65  draft  gear 
at  g,  if  the  constant  dividend  is  381? 

Solution. —  381  4-  65  =  5.86,  draft.  Ans. 

To  find  the  draft  gear  when  the  constant  dividend  and 
draft  are  known: 


Rule  III. — Divide  the  constant  dividend  by  the  draft  desired. 

Example  3. — What  draft  gear  will  be  required  to  produce  a  draft 
of  5.86  if  the  constant  dividend  is  381? 


Solution. —  381  4-  5.86  =  65,  draft  gear.  Ans. 

Example  4. — Figure  the  constant  for  Fig.  4,  using  the  same  train 
of  gears  as  in  the  previous  examples  but  considering  the  gear  f  as  the 
draft  change  gear. 

11  X  1  X  70 


Solution. — 


=  .0598,  constant  factor.  Ans. 


9  X  22  X  65 

To  find  the  draft  when  the  constant  factor  and  draft  gear 
are  known: 


Rule  IV. — Multiply  the  constant  factor  by  the  draft  gear. 

Example  5. — What  is  the  total  draft  for  Fig.  4,  considering  f  as 
the  draft  gear,  if  the  constant  factor  is  .0598,  a  98-tooth  gear  being 
used  at  /? 

Solution. —  .0598  X  98  =  5.86,  draft.  Ans. 

To  find  the  draft  gear  when  the  constant  factor  and  draft 
are  known: 

Rule  V. — Divide  the  draft  by  the  constant  factor. 

Example  6. — What  draft  gear  will  be  required  at  f,  Fig.  4,  to 
produce  a  draft  of  5.86  if  the  constant  factor  with  f  considered  as  the 
change  gear  is  .0598? 

Solution. —  5.86  4-  .0598  =  97.99  =  98,  draft  gear.  Ans. 


30 


DRAFT  CALCULATIONS 


§15 


From  the  examples  previously  given  for  finding  the  draft 
gear,  it  will  be  noticed  that  the  solution  does  not  always  give 
an  exact  number  of  teeth  for  the  change  gear.  In  such  cases 
the  nearest  number  is  used.  For  example,  if  the  solution  of 
a  draft  calculation  should  show  that  a  64.84  draft  gear  was 
required,  then  a  65  gear  would  in  all  probability  be  placed  on 
the  machine,  and  even  if  the  calculation  should  show  that  a 
64.52  draft  gear  was  required,  a  65  gear  would  be  used, 
except  in  cases  where  extreme  accuracy  was  desired.  Under 
these  circumstances  either  the  back-roll  gear  or  the  crown 
gear  would  be  changed.  When  the  crown,  or  back-roll,  gear 
is  changed,  it  is  generally  considered  that  one  tooth  in  the 
draft  gear  is  equal  to  two  teeth  in  the  crown,  or  back-roll, 
gear.  This  allowance  is  not  mathematically  correct,  but  it  is 
near  enough  for  practical  purposes  and  is  the  basis  generally 
adopted  in  the  mill.  For  example,  a  draft  gear  figures  42-2 
with  a  60  back-roll  gear.  A  42^-  draft  gear  cannot  be  used, 
so  a  42  draft  gear  and  a  59  back-roll  gear,  or  a  43  draft  and  a 
61  back-roll  gear  would  probably  be  used. 


DOUBLING 

23.  When  calculating  what  effect  the  draft  of  a  machine 
will  have  on  the  weight  of  the  sliver  or  roving,  it  is  always 
necessary  to  take  into  consideration  the  number  of  ends  that 
are  to  be  drawn  into  one.  For  example,  six  ends  of  roving 
are  run  into  one  in  a  certain  machine  that  has  a  draft  of  6; 
consequently,  each  end  of  roving  will  have  to  be  drawn  out 
to  one-sixth  of  its  former  weight;  but  since  there  are  six  ends 
running  into  one/*  then  the  weight  per  yard  of  the  sliver 
delivered  will  be  the  same  as  the  weight  per  yard  of  a  single 
sliver  put  up  at  the  back.  Therefore,  if  six  slivers,  each 
weighing  65  grains  to  the  yard,  are  run  through  a  machine 
having  a  draft  of  6,  the  sliver  that  comes  out  at  the  front  will 
have  the  same  weight;  that  is,  65  grains.  Hence,  when 
figuring  the  weight  of  cotton  in  connection  with  the  draft  of  a 
machine,  it  is  always  necessary  to  take  into  consideration  the 


15 


DRAFT  CALCULATIONS 


31 


number  of  ends  that  are  placed  at  the  back  and  run  into  a 
single  end  at  the  front. 

To  find  the  weight  of  a  sliver  or  roving  produced  by  a 
machine  when  the  draft  of  the  machine  and  the  number  and 
weight  of  the  ends  put  up  at  the  back  are  known: 


Rule  I. — Multiply  the  weight  per  yard  of  the  roving  or  sliver 
at  the  back  by  the  number  of  ends  run  into  one  at  the  back  and 
divide  this  product  by  the  draft  of  the  machine. 


Example  1. — A  certain  machine  has  a  draft  of  6.  What  will  be  the 
weight  of  finished  sliver  if  six  ends,  each  weighing  65  grains,  are  put 
up  at  the  back  and  run  into  one? 


Solution. — 


65  X  6 
6 


=  65  gr.  per  yd. 


Ans. 


To  find  the  draft  of  a  machine  when  the  number  of  ends 
at  the  back,  the  weight  of  the  sliver  at  the  back,  and  the 
weight  of  the  sliver  delivered  are  known: 


Rule  II. — Multiply  the  weight  per  yard  of  the  sliver  at  the 
back  by  the  number  of  ends  run  into  one  at  the  back  and  divide 
this  product  by  the  weight  per  yard  of  the  sliver  delivered  at  the 
front. 

Example  2. — At  a  certain  machine  six  ends  of  sliver,  each  weighing 
65  grains,  are  put  up  at  the  back  and  run  into  one.  What  is  the  draft 
of  the  machine  if  1  yard  of  finished  sliver  weighs  65  grains? 

Solution.—  =  6,  draft.  Ans. 


24.  The  basis  for  the  draft  calculations  given  in  the 
preceding  examples  has  been  the  weight  of  the  sliver, 
expressed  in  grains  per  yard.  Frequently,  however,  calcula¬ 
tions  must  be  made,  using  as  a  basis  the  hank  of  the  roving, 
the  number  of  the  yarn,  or  both  of  these  items,  thus  requiring 
the  use  of  a  rule  exactly  the  reverse  of  the  one  used  when 
figuring  on  the  basis  of  grains  per  yard.  This  is  for  the 
reason  that,  when  indicating  the  weight  of  a  sliver  or  a 
roving  by  its  weight  per  yard,  a  greater  number  is  used  to 
indicate  a  heavier  sliver  or  roving,  and  a  smaller  number  to 
indicate  a  finer  sliver  or  roving;  but,  when  indicating  the 
weight  of  sliver  or  roving  by  the  hank,  just  the  reverse  is 


32 


DRAFT  CALCULATIONS 


§15 


the  case,  as  a  smaller  number  is  used  to  indicate  a  heavier 
sliver  or  roving,  and  a  larger  number  to  indicate  a  finer 
sliver  or  roving. 

The  following  rules  and  examples  will  be  found  to  apply 
to  draft  calculations  when  the  weight  of  the  sliver  or  roving 
is  expressed  in  hanks. 

To  find  the  hank  of  a  roving  made  by  a  machine  when  the 
draft  of  the  machine  and  the  number  and  hank' of  the  ends 
put  up  at  the  back  are  known: 

Rule  I . — Multiply  the  hank  of  the  roving  at  the  back  by  the 
draft  of  the  machine  a?id  divide  this  product  by  the  number  of 
ends  put  tip  at  the  back. 

Example  1. — A  certain  machine  has  a  draft  of  6.  What  will  be  the 
hank  of  finished  roving  if  two  ends,  each  3-hank,  are  put  up  at  the 
back  and  run  into  one  at  the  front? 

3X6 

Solution. —  .  — =  9-hank.  Ans. 

•u 

To  find  the  draft  of  a  machine  when  the  number  of  ends  at 
the  back,  the  hank  of  the  roving  at  the  back,  and  the  hank 
of  the  roving  delivered  are  known: 

Rule  II. — Multiply  the  hank  of  the  roving  delivered  by  the 
number  of  ends  put  up  at  the  back  and  divide  by  the  hank  of  the 
roving  used  at  the  back. 

Example  2. — At  a  certain  machine  two  ends  of  roving,  each  3-hank, 
are  put  up  and  run  into  one  at  the  front.  What  is  the  draft  of  the 
machine  if  the  finished  roving  is  9-hank? 

9X2 

Solution. —  =  6,  draft.  Ans. 

Attention  has  previously  been  drawn,  in  Arts.  19  and  20, 
to  the  fact  that  certain  rules  are  largely  modified  according 
to  whether  the  draft  change  gear  is  a  driver  or  a  driven, 
while  it  is  here  shown  that  certain  other  rules  are  affected  by 
the  method  of  expressing  the  weight  of  the  sliver  or  roving. 
In  practical  draft  calculations  occurring  in  cotton-mill  work 
it  will  frequently  be  found  necessary  to  modify  or  change 
the  rules  that  have  been  given,  in  order  to  meet  special 


§15 


DRAFT  CALCULATIONS 


33 


circumstances,  since  rules  for  draft  are  affected  not  only  by 
the  points  just  referred  to,  but  also  by  numerous  other  con¬ 
ditions  that  may  arise  with  special  arrangements.  This 
emphasizes  the  necessity  for  a  careful  study  of  the  funda¬ 
mental  principles  of  drafting  and  draft  calculations.  When 
once  these  principles  are  thoroughly  understood,  it  is  pos¬ 
sible  to  dispense  entirely  with  reference  to  rules,  as  a  perfect 
knowledge  of  the  subject  enables  the  person  making  the 
calculations  to  know  intuitively  whether  each  different  factor 
of  the  example  is  a  multiplier  or  divider,  through  knowing 
the  cause  and  effect  of  an  alteration  in  each  factor. 


READING  TEXTILE 
DRAWINGS 


REPRESENTATION  OF  OBJECTS 


INTRODUCTION 

Note. — Many  students  fail  to  understand  fully  the  various  drawings 
contained  in  their  Course,  and  in  consequence  do  not  obtain  as 
thorough  a  knowledge  of  the  machines  and  mechanisms  incident  to 
the  various  processes  dealt  with  as  would  be  possible  if  the  drawings 
were  clearly  comprehended.  As  it  is  impossible  to  instruct  the  student 
in  the  theory  and  practice  of  the  many  processes  occurring  in  the  vari¬ 
ous  branches  of  the  textile  industry  without  making  use  of  drawings 
to  illustrate  the  instruction  and  enhance  its  value,  this  short  Section 
on  reading  drawings  has  been  prepared.  It  is  not  intended  to  teach 
the  student  how  to  make  a  drawing  or  how  to  read  the  working  draw¬ 
ings  used  in  modern  shop  practice;  the  only  use  of  this  Section  is  to 
enable  the  student  to  understand  clearly  the  drawings  of  the  machines 
and  mechanisms  contained  in  his  Course,  so  that  they  will  be  as  valu¬ 
able  to  him  as  the  text  of  the  lessons.  The  descriptions  given  and 
methods  dealt  with  in  this  Section  are  intended  to  apply  only  to  the 
Instruction  Papers  of  the  International  Correspondence  Schools,  and 
hence  should  not  be  considered  as  of  universal  application. 

1.  A  drawing  is  a  representation,  by  means  of  lines,  of 
any  object  whatsoever,  as  for  instance  a  machine  or  a  part 
of  a  machine.  Drawing  has  been  spoken  of  as  the  “language 
of  the  engineer.”  In  addition  it  is  also  a  universal  lan¬ 
guage.  A  drawing  made  in  France  or  Germany  can  be  inter¬ 
preted  as  readily  in  the  United  States  of  America  as  can 
a  drawing  made  in  one  of  the  drafting  rooms  of  the  latter 
country,  except,  of  course,  that  any  notes  or  other  written 
matter  on  the  drawing  would  be  in  a  foreign  language, 
and  the  units  of  measurement  in  a  different  system.  The 


For  notice  of  copyright,  see  page  immediately  following  the  title  page 

2  91 


2 


READING  TEXTILE  DRAWINGS 


§91 


drawing  itself,  however,  would  be  quite  as  intelligible.  A 
drawing,  while  it  is  a  representation  of  an  object,  is  not  a 
photograph  of  that  object,  although  certain  types  of  drawings 
give  to  the  eye  an  impression  similar  to  that  given  by  a 
photographic  reproduction  of  the  object.  A  drawing,  there¬ 
fore,  may  be  defined  as  a  conventional  representation  of  an 
object  by  means  of  lines.  Although  drawings  are  of  especial 
importance  to  an  engineer,  they  are  of  inestimable  value  to 
those  engaged  in  the  various  branches  of  the  textile  industry. 


KINDS  OF  DRAWINGS 


MECHANICAL  AND  FREEHAND  DRAWINGS 

2.  In  general  there  are  two  fundamental  kinds  of  draw¬ 
ings — freehand  drawings  and  mechanical  drawings.  Mechan¬ 
ical  drawings  are  made  with  the  aid  of  various  instruments 
such  as  T  squares,  triangles,  compasses,  etc.  Freehand 
drawings  are  made  without  such  aids  to  accuracy,  and  the 
draftsman  depends  solely  on  his  practiced  eye  in  determin¬ 
ing  the  positions  and  lengths  of  lines  and  on  his  skilled  hand 
in  drawing  them.  While  the  ability  to  make  good  freehand 
drawings  is  of  value  mainly  to  the  artist,  it  is  likewise  a 
valuable  attribute  of  the  artisan,  particularly  in  cases  of 
emergency  where  a  good  sketch  may  often  take  the  place  of 
a  mechanical  drawing.  In  fact,  drawings  executed  accord¬ 
ing  to  the  rules  of  mechanical  drawing  are  by  common 
consent  classified  as  mechanical  drawings  whether  made  free¬ 
hand  or  not.  Mechanical  drawings  serve  as  the  connect¬ 
ing  link  between  the  machine  designer  and  the  artisan, 
giving  the  latter  detailed  information  as  to  the  construction 
of  a  machine.  In  addition,  they  are  of  value  in  illustrating 
the  construction  and  operation  of  machines  and  mechanisms, 
in  which  case  dimensions  and  other  specifications  that  accom¬ 
pany  drawings  from  which  a  machine  is  to  be  constructed 
are  generally  omitted. 

3.  When  an  object  is  represented  by  means  of  a  mechan¬ 
ical  drawing,  the  appearance  of  the  latter  differs  according  to 


91 


READING  TEXTILE  DRAWINGS 


3 


the  position  assumed  by  the  draftsman  in  viewing  the  object. 
Any  one  of  the  drawings  made  by  assuming  an  object  to  be 
seen  from  different  positions  is  called  a  view ,  but  distinction 
is  made  between  the  several  views  according  to  which  side 
of  the  object  they  represent.  The  more  important  ones  are: 
front  view ,  side  view ,  and  plan  view.  Then  there  are  imagi¬ 
nary  views  in  which  it  is  supposed  that  some  part  of  the  object 
is  cut  away  in  order  to  show  its  interior  construction;  these 
are  termed  sectional  views.  Views  that  represent  a  machine 
or  mechanism  with  every  part  in  position  are  collectively 
termed  an  assembly  drawing.  If,  on  the  contrary,  various 
views,  each  representing  a  single  element  of  a  machine  are 
shown,  they  constitute  a  detail  drawing. 

4.  Other  views  are:  perspective  and  diagrammatic  views. 
These,  while  made  by  the  aid  of  drawing  instruments,  are 
generally  not  classed  as  mechanical  drawings,  although  they 
will  be  so  considered  here. 


MECHANICAL  DRAWINGS 


THE  VARIOUS  VIEWS  AND  THEIR  ARRANGEMENT 


FUNDAMENTAL  PRINCIPLES 

5.  When  a  mechanical  drawing  contains  more  than  one 
view  of  an  object,  such  views  -are  arranged  according  to 
certain  definite  rules.  If  so  arranged,  the  position  of  each 
view  will  at  once  indicate  which  side  of  the  object  it  is 
intended  to  represent,  thus  obviating  the  necessity  of  adding 
any  special  information  as  to  the  point  of  view  from  which  it 
was  taken. 

To  explain  the  basis  of  these  rules,  let  it  be  assumed  that 
several  views  are  to  be  made  of  the  object  shown  in  Fig.  1. 
This  object  is  rectangular  in  shape,  has  a  knob  g  on  its  front 
end,  a  square  boss  h  on  its  right  side,  a  semicircular  piece  i 
with  a  hole  through  it  on  its  rear  end,  a  hexagonal  boss  j  on 


4 


READING  TEXTILE  DRAWINGS 


91 


its  left  side,  and  two  cylindrical  studs  k,  kx  on  its  top.  The 
pieces  j  and  i  and  the  outlines  of  the  block  shown  in  dotted 
lines  are  those  parts  of  the  object  that  cannot  be  seen  in  its 
present  position.  A  glass  case,  indicated  by  dotted  lines  in 
Fig.  2,  is  supposed  to  enclose  the  object,  and  as  the  assump¬ 


tion  is  that  each  of  its  sides  is  hinged  to  the  top,  it  is  possible 
to  swing  them  into  the  positions  indicated  by  full  lines.  It 
is  further  assumed  that  the  exterior  surfaces  of  the  case  are 
so  prepared  that  they  can  be  drawn  on  by  means  of  a  pencil. 
The  draftsman  is  then  supposed  to  place  himself,  succes¬ 


sively,  in  front  of  each  side,  also  above  the  top,  and  to  trace 
the  outlines  of  the  object  on  the  plate  intervening  between 
himself  and  the  object.  When  all  the  sides  are  swung  on 
their  hinges  until  level  with  the  top,  as  indicated  by  the 
full  lines  in  Fig.  2,  it  will  be  seen  that  the  draftsman  has 


Fig.  3 


6 


READING  TEXTILE  DRAWINGS 


§91 


produced  a  series  of  views  grouped  around  a  central  view. 
These  views  are  also  shown  on  their  respective  plates,  but  on 
a  larger  scale,  in  Fig.  3,  and  are  arranged  according  to  the 
rules  of  mechanical  drawing.  In  order  to  complete  the  series 
it  should  be  supposed  that  there  is  an  additional  glass  plate 
under  the  object  and  that  this  plate  is  hinged  to  one  of  the 
other  plates,  in  this  case  to  plate  a.  It  is  swung  into  posi¬ 
tion  with  the  rest  of  the  plates  in  Fig.  3  and  is  represented 
by  the  plate  /  shown  in  dotted  lines.  The  views  here  given 
are  classed  under  two  heads:  elevations ,  or  side  views ,  and 
plan  viezvs,  or  simply  plans. 

Note. — In  lettering  Fig.  3  some  deviation  has  been  made  from  the 
rules  in  mechanical  drawing  in  order  that  the  elementary  character  of 
this  Section  may  be  maintained.  According  to  the  rules,  the  tops  of 
the  reference  letters  and  figures,  found  in  the  different  views,  should 
all  point  toward  the  side  considered  as  the  top  side  of  the  drawing. 
The  advantage  of  this  arrangement  is  that  the  letters  in  all  the  views 
may  be  read  without  changing  the  position  of  the  drawing.  It  is 
noticed,  however,  that  the  lettering  on  Fig.  3  does  not  follow  this  rule, 
that  in  fact  the  letters  point  in  four  directions  and  that  therefore  the 
drawing  must  be  turned  into  four  positions  before  the  letters  in  all  the 
views  maybe  read.  It  was  thought  to  be  less  confusing  to  the  student 
if  he  were  enabled  to  examine  the  various  views,  when  they  were  held 
in  such  a  position  that  the  object,  represented  by  them,  were  resting 
on  its  natural  base,  and  the  letters  were  arranged  so  as  to  indicate 
this  position.  When  a  single  view  is  made  of  an  object  it  will,  in 
general,  occupy  a  similar  position,  but  in  a  collection  of  views,  as  in 
Fig.  3,  some  of  the  views  will  show  the  object  standing  on  one  of  its 
sides  or  on  its  top.  This  cannot  be  otherwise  and  does  in  no  manner 
affect  the  correctness  of  the  drawing. 

It  is  understood  that  the  rectangles  surrounding  the  various  views 
in  Fig.  3,  and  which  are  supposed  to  represent  plates  on  which  the 
views  are  drawn,  are  omitted  in  regular  mechanical  drawings.  The 
views  themselves,  however,  remain  in  the  positions  indicated. 


ELEVATIONS,  OR  SIDE  VIEWS 

6.  Elevations  are  views  of  the  sides  or  ends  of  an 
object.  Thus,  the  object  shown  in  Fig.  1  may  have  four  eleva¬ 
tions,  or  side  views:  a  front  elevation ,  which  shows  the 
appearance  of  the  object  when  seen  from  the  direction  indi¬ 
cated  by  the  arrow  a ,  Fig.  1;  a  right-side  elevation ,  which 
shows  the  appearance  of  the  object  as  seen  from  the  direc¬ 
tion  of  the  arrow  b\  a  rear  elevation ,  which  shows  the  appear¬ 
ance  of  the  object  as  seen  from  the  direction  of  the  arrow  c; 


§91 


READING  TEXTILE  DRAWINGS 


7 


and  the  left-side  elevation ,  which  shows  the  appearance  of  the 
object  as  seen  from  the  direction  of  the  arrow  d. 

In  Fig.  3,  these  elevations,  or  side  views,  are  shown  in 
views  (a),  (b),  (c) ,  and  (d) ,  which 'are  lettered  in  accordance 
with  the  arrows  indicated  in  Fig.  1.  The  front  elevation  is 
shown  in  view  (a),  the  right-side  elevation  in  (b) ,  the  rear 
elevation  in  (e) ,  and  the  left-side  elevation  in  (d).  Each 
elevation  shows  only  one  side  of  the  object  and,  therefore, 
if  considered  alone,  gives  no  indication  of  the  depth  or  thick¬ 
ness  of  same. 


PLANS 

7.  In  addition  to  the  elevations  of  an  object,  plans  may 
also  be  given.  A  plan  is  a  view  of  an  object  as  seen  from 
a  point  directly  above  or  below;  that  is,  it  is  a  view  of  the 
top  or  bottom  of  an  object.  Two  plan  views  may  therefore 
be  shown  of  the  object  in  Fig.  1,  one  as  seen  in  the  direction 
of  the  arrow  e  and  the  other  in  that  of  the  arrow  /.  The  first 
is  termed  a  top  plan ,  and  the  latter  an  inverted  plan.  Inverted 
plans  are  rarely  drawn  and  when  the  term  plan  is  used  a  top 
plan  is  generally  inferred.  In  Fig.  3,  a  top  plan  is  shown  on 
plate  {e)  and  an  inverted  plan  on  plate  (/). 

8.  Fig.  3  shows  a  complete  set  of  views  of  the  object 
under  consideration,  but  it  is  seldom  that  so  many  are  given; 
the  number  depends  to  a  great  extent  on  the  shape  of  the 
object  and  the  necessity  of  making  its  details  perfectly  clear. 
For  instance,  in  the  case  of  a  cube  only  two  views  are 
required,  such  as  two  side  elevations.  It  should  also  be 
understood  that  it  is  not  necessary  always  to  group  the  views 
after  the  manner  of  Fig.  3;  that  is,  around  the  plan  view. 
The  view  on  any  one  side  of  the  case  outlined  in  Fig.  2  may 
be  used  as  the  main,  or  central,  view,  but  care  must  be  taken 
to  arrange  the  additional  views  in  their  proper  positions. 

9.  When  fewer  views  of  an  object  than  shown  in  Fig.  3 
are  given,  their  arrangement,  while  governed  by  the  same 
laws  as  were  previously  laid  down,  often  differs  somewhat. 
Thus,  suppose  that  front  and  side  elevations  are  to  be  drawn 


8 


READING  TEXTILE  DRAWINGS 


§91 


of  the  irregular  object  A  shown  in  Fig.  4.  The  front  view  a 
in  this  case  may  be  selected  as  the  central  view;  but  then 
it  is  important  that  the  right-side  elevation  be  placed  to  the 
right,  and  the  left-side  elevation  to  the  left,  in  order  to  retain 
the  original  position  of  the  plates  b  and  c,  as  indicated  by  the 
dotted  lines.  When  these  plates  are  swung  into  position  so  as 
to  constitute  a  continuation  of  the  plate  a ,  the  corresponding 
views  will  be  correctly  arranged,  and  it  will  be  further  noted 
that  each  view  shows  the  object  resting  on  its  natural  base. 

10.  In  case  it  should  be  required  to  add  a  rear  view  of  the 
object  in  Fig.  4,  this  view  would  be  placed  alongside  view  b 


or  c.  If  a  top  view  were  added,  it  would  be  above  view  a , 
while  an  inverted  plan  would  be  placed  below.  The  positions 
of  the  views  of  an  object  relative  to  one  another  will  there¬ 
fore  always  indicate  from  which  side  the  view  is  taken. 

11.  As  further  illustrations  of  various  views  Figs.  5  and  6 
are  given.  Fig.  5  (a)  is  a  front  elevation  of  the  shedding 
mechanism  of  a  5-harness  side-cam  loom  and  (b)  a  right-side 
elevation  of  the  same.  Being  on  the  right  side  of  Fig.  5  (a) , 
it  is  at  once  understood  that  it  is  a  right-side  view  of  same. 
Fig.  6  is  a  plan  view  of  a  cotton  card,  showing  the  location 
of  the  lap  roll,  feed-roll,  licker,  cylinder,  doffer,  etc.;  it 
illustrates  also  the  method  of  driving  the  various  parts. 


miiiii  . . 


Fig. 


Fig.  6 


§91 


READING  TEXTILE  DRAWINGS 


11 


SECTIONAL  VIEWS 

12.  Sectional  views  are  those  that  show  the  interior 
construction  of  an  object,  which  is  assumed  to  be  cut  or 
broken  along-  a  certain  plane  or  through  certain  parts.  By 
the  word  plane  is  meant  a  perfectly  uniform,  flat  surface. 


theoretically  without  thickness,  which  in  the  case  of  sectional 
views  is  assumed  to  be  a  cutting  surface,  or  plane.  For 
instance,  suppose  that  a  longitudinal  section  of  the  object 
shown  in  Fig.  1  was  desired;  then  the  object  would  be 


Fig.  8 

assumed  to  be  cut,  as  shown  in  Fig.  7,  by  a  plane  /  no  m  and 
one  of  the  halves  removed.  The  sectional  drawing  would 
be  made  to  show  the  appearance  of  the  cut  surface  in  eleva¬ 
tion,  as  shown  in  Fig.  8.  The  cutting  plane  on  which  a 


12 


READING  TEXTILE  DRAWINGS 


§91 


section  is  taken  may  be  indicated  by  a  dot-and-dash  line  on 
another  view  of  the  object.  In  the  case  of  a  cutting  plane 
passing  longitudinally  through  an  object,  as  shown  in  Fig.  7, 
it  could  be  indicated  on  either  end  view  of  the  object,  as 
shown  by  the  line  l  m,  Fig.  3  {a) ,  or  n  o ,  view  (c) ,  or  on  the 
top  plan  view,  as  indicated  by  the  line  n  l,  view  (<?),  or  on 
the  inverted  plan  view,  as  shown  by  the  line  o  m,  view  (/). 
In  these  cases,  it  could  be  stated  that  Fig.  8  was  a  section 
taken  through  l  m,  Fig.  3  ( a );  n  o,  view  ( c );  nl ,  view  {e)-y 
or  o  m,  view  (/) . 

13.  Wherever  an  object  or  any  part  of  an  object  that  is 
supposed  to  be  cut  or  broken  away  is  shown  in  section,  the 
material  of  which  that  object  is  made  is  usually  indicated 


Fir,  9 


by  certain  lines  or  combinations  of  lines.  Unfortunately 
there  is  no  universally  adopted  standard  for  thus  indicating 
different  materials.  For  instance,  a  certain  combination  of 
lines  will  indicate  that  the  material  is  cast  iron  if  drawn  in 
one  drafting  room,  while  if  drawn  in  another,  this  same 
combination  may  have  been  adopted  to  represent  brass. 

14.  The  most  commonly  used  combination  of  lines  for 
different  materials  is  shown  in  Fig.  9.  Steel  of  all  kinds  is 
indicated  as  shown  at  A\  B  shows  the  style  of  sectioning 
employed  for  wrought  iron.  Cast  iron  is  usually  sectioned 
as  shown  at  C;  thus,  Fig.  8  shows  that  the  object  shown  in 
Fig.  1  is  one  solid  piece  of  cast  iron  through  that  portion 
where  the  section  is  taken.  Brass  and  other  similar  copper 


§91 


READING  TEXTILE  DRAWINGS 


13 


alloys  are  sectioned  as  shown  at  D,  Fig.  9.  For  lead,  Babbitt 
metal,  and  similar  soft  materials,  the  sectioning  shown  at 
E  is  extensively  used.  Wood,  when  cut  across  the  grain,  is 
usually  sectioned  as  shown  in  the  upper  half  of  F,  and  when 
cut  along  the  grain  as  shown  in  the  lower  half.  Wood  is 
also  frequently  indicated  on  a  drawing  by  section  lines,  even 
when  it  is  not  a  section.  Glass  and  stone,  when  in  section,, 
are  often  indicated  in  the  manner  shown  by  the  upper  half  of  G; 
when  not  in  section,  they  are  frequently  drawn  as  shown  in 
the  lower  half  of  that  view.  Sometimes  the  method  shown 
in  the  lower  half  of  view  G  is  used  to  indicate  polished  sur¬ 
faces,  as  for  instance  blades  of  knives,  etc.  Concrete  may 
be  indicated  as  in  H\  I  is  a  common  representation  of 
leather.  Rubber  and  wood  fiber  are  sectioned  as  in  J,  fire¬ 
brick  as  in  A",  and  water  as  in  L. 


As  an  illustration  of  the  use  of  section  lining,  suppose  that 
the  object  shown  in  Figs.  1  and  8,  instead  of  being  one  solid 
piece  of  cast  iron,  was  made  with  a  body  of  cast  iron  but 
with  the  pieces  k ,  k x  of  wrought  iron,  g  of  brass,  and  i  of 
steel.  The  section  of  this  object  would  then  appear  as  in 
Fig.  10  instead  of  as  in  Fig.  8,  the  different  kinds  of  section 
lining  showing  the  material  from  which  each  part  of  the 
object  was  made. 

15.  In  the  modern  drawing  office,  it  is  customary  to  use 
the  type  of  cross-sectioning  shown  at  C,  Fig.  9,  for  all  mate¬ 
rials,  unless  it  is  deemed  advisable  on  complicated  drawings 
to  designate  kinds  of  materials  by  character  of  cross-section 
lining,  in  which  case  the  standard  sections  shown  are  used. 
In  these  Courses,  the  same  method  is  followed,  differentiation 
between  materials  being  made  only  in  special  cases  where,  for 
instance,  wood  and  fibrous  materials  are  shown  in  section. 


14 


READING  TEXTILE  DRAWINGS 


§91 


16.  Sections  of  material  that  are  too  small  on  a  drawing 
to  be  conveniently  sectioned,  or  when  it  is  desired  to  make 
the  section  very  prominent,  are  often  blackened  in,  as  shown 
in  Fig.  11.  In  order  to  separate  different  pieces,  a  white 
line  is  then  usually  left  between  them.  Sometimes  sectional 


Fig.  11 

views  are  shown  on  a  drawing  that  do  not  represent  a  com¬ 
plete  division  of  an  object  by  the  cutting  plane;  that  is,  they 
are  only  partial  sections.  This  is  usually  done  for  disclosing 
some  special  feature  of  the  object.  In  such  a  case  a  part  of 
the  object  is  broken  away  by  a  freehand,  wavy  line  drawn 
around  the  partial  section  and  the  broken  part  shown  in  sec¬ 
tion.  For  instance,  in 
Fig.  12  one  of  the  jaws 
of  the  rod  there  shown 
has  been  partly  broken 
away  at  a  in  order 
to  show  clearly  that  a 
dowel  pin  has  been  in¬ 
serted  for  the  purpose 
of  keeping  the  bolt 
from  turning. 

17.  On  many  sec¬ 
tional  views,  it  will  be 
noticed  that  the  sec¬ 
tion  lines  do  not  run  in  the  same  direction.  This  invari¬ 
ably  means  that  there  is  more  than  one  piece  in  the  section 
given;  thus,  referring  to  Fig.  13,  it  will  be  seen  that  the  sec¬ 
tion  lining  shown  at  b ,  b  is  at  right  angles  to  the  other  sec¬ 
tion  lining.  It  is  a  general  rule  among  draftsmen  that  all 
parts  of  the  same  piece  shown  in  section  must  be  section-lined 


§91 


READING  TEXTILE  DRAWINGS 


15 


in  the  same  direction,  irrespective  of  the  continuity  of  the 
section.  Thus,  referring  again  to  Fig.  13,  the  fact  that  all 
section  lining  marked  A  is  in  the  same  direction  immediately 
establishes  the  fact  that  this  part 
of  the  view  is  a  section  of  the 
same  piece;  likewise,  since  all 
the  sectioning  marked  b,  b  runs 
in  the  one  direction,  it  follows 
that  b,  b  are  sectional  views  of 
one  piece,  which  is  a  separate 
piece  from  the  piece  A.  It  will  also  be  seen  in  Fig.  10  that  the 
section  lines  on  the  separate  parts  k,  kx,  g,  i ,  run  in  a  direction 
at  right  angles  to  those  on  the  body  of  the  object. 

18.  Longitudinal  sections  are  those  taken  lengthwise 
through  an  object  or  parallel  to  its  known  axis,  if  the  object 
has  an  axis.  A  cross-section  is  one  that  is  taken  at  any 
angle,  usually  at  right  angles,  to  the  direction  in  which  a 
longitudinal  section  would  be  taken. 

Fig.  14  (a)  shows  the  longitudinal  section  of  a  hollow 
shaft,  in  the  end  of  which  a  rod  is  inserted  and  held  in  place 


a 

1 


,  b 

(b)  (, a ) 

Fig.  14 


by  a  tapered  cotter  pin.  Fig.  14  (b)  shows  a  cross-section 
of  this  arrangement  taken  through  the  line  a  b,  Fig.  14  (a). 
It  will  be  noted  in  this  case  that  the  tapered  pin  is  not  shown 
in  section  but  is  shown  in  full.  In  cases  like  this  where  a 
pin  or  key  or  other  object  of  similar  construction  is  shown  in 
a  section,  this  part  is  usually  shown  full,  because  it  makes 
the  drawing  easier  to  read  and  also  saves  considerable  time 
in  making  the  drawing. 


16 


READING  TEXTILE  DRAWINGS 


§91 


19.  Draftsmen  occasionally  make  use  of  conventional 
methods  of  taking  sections  for  the  purpose  of  better  convey¬ 
ing  the  meaning  of  the  section  to  the  mind.  While  there  is 
no  universal  rule  followed,  a  few  examples  will  be  suggest¬ 
ive  of  others,  and  the  points  brought  out  will  allow  other 
instances  of  conventional  sectioning  to  be  recognized. 

Fig.  15  (a)  and  (/;)  shows  a  front  and  a  sectional  view  of 
a  pulley.  The  sectional  view  is  taken  along  the  line  A  D , 
which  passes  through  two  of  the  arms.  If  the  sectional 
view  is  drawn  strictly  according  to  the  rules,  it  will  appear 
as  in  view  (b);  that  is,  the  arms  will  be  sectioned,  since  they 


Fig.  15 


are  cut  by  the  cutting  plane.  Now,  if  the  sectional  view 
(i b)  be  considered  by  itself,  it  will  convey  the  idea  to  most 
persons  that  there  is  a  solid  web  between  the  rim  and  the 
hub  of  the  pulley.  While  reference  to  the  front  view  will 
immediately  dispel  this  impression,  the  fact  remains  that, 
without  a  second  view,  the  conclusion  that  a  solid  web 
exists  would  be  justifiable.  Therefore,  in  a  case  like  the 
one  shown,  in  order  to  convey  the  idea  better,  the  drafts¬ 
man  will  often  make  the  section  on  the  assumption  that 
the  arms  are  just  back  of  the  cutting  plane,  so  that  they 
will  appear  in  full  in  the  sectional  view,  or  as  shown  in 
view  (c) . 


§91 


READING  TEXTILE  DRAWINGS 


17 


DETAIL  AND  ASSEMBLY 
DRAWINGS 

20.  Detail  draw- 
♦  ings  are  often  used  in 
connection  with  other 
drawings  to  show  clearly 
objects  or  parts  of  ob¬ 
jects  that  are  not  clearly 
shown  in  the  other 
views.  For  instance, 
Fig.  17  shows  a  view  of 
a  loom  temple,  while 
Fig.  18  shows  the  de- 


(a) 


Referring  now  to  Fig.  15  (a),  a  section  is  shown  at  a. 
This  is  a  conventionalism  frequently  employed  to  indicate  the 
shape  of  the  cross-section  of  the  part  on  which  it  is  placed. 
In  this  case,  it  shows  that  the  arms  are  elliptic  in  cross- 
section.  As  a  further 
illustration  of  sectional 
drawings,  Fig.  16  is 
given.  Fig.  16  ( a )  is 
an  elevation  of  a  spin¬ 
ning  spindle,  while 
Fig.  16  (b)  is  a  longi¬ 
tudinal  section,  showing 
the  interior  construc¬ 
tion  and  arrangement 
of  the  various  parts  of 
the  spindle. 


0>) 


Fig.  16 


tails  of  its  various 
parts.  Fig.  17  would 
be  known  as  an  assembly  drawing,  and  Fig.  18  as  the  detail 
drawings. 

It  frequently  happens  that  in  large  assembly  drawings 
certain  parts  are  necessarily  so  small  as  to  render  them 
indistinct;  in  such  cases,  detail  drawings  of  these  parts  are 


18 


READING  TEXTILE  DRAWINGS 


§91 


often  given.  These  detail  drawings  may  not  necessarily 
be  drawings  of  a  single  part,  but  may  be  enlarged  draw¬ 
ings  of  a  group  of 
parts  or  of  the  assem¬ 
bled  parts  of  some  par¬ 
ticular  mechanism. 


PERSPECTIVE  VIEWS 

21.  A  perspec¬ 
tive  view  may  be  said 
to  combine  several 
views  of  an  object  into 
one.  For  this  reason 
it  gives  at  one  glance 
a  correct  idea  as  to  the 
shape  of  an  object,  while  in  a  regular  mechanical  drawing  it 
may  be  necessary  to  study  several  views  before  the  mind 


h 

© 

© 

i 

« 

o2 

J 

too 

w 

Fig.  18 


will  be  able  to  imagine  its  true  shape.  A  perspective  view 
is  drawn  so  as  to  represent  the  object  as  it  will  appear  to  the 
eye  when  viewed  under  ordinary  conditions,  while  the  views 


\ 


§91 


READING  TEXTILE  DRAWINGS 


19 


given  in  mechanical  drawings  are,  for  the  sake  of  convenience, 
drawn  under  assumed  conditions,  as  previously  described. 


Fig.  1  is  an  example  of  a  perspective  view.  It  may  be 
said  in  this  instance  to  combine  a  front  elevation,  a  right- 
side  elevation,  and  a  plan  view  in  one,  and  gives  at  once 
a  good  idea  of  the  shape  of  the  object. 


20 


READING  TEXTILE  DRAWINGS 


§91 


As  a  further  illustration  of  perspective  views,  Fig.  19  is 
given.  This  represents  a  loom  as  it  appears  when  viewed 
from  a  point  in  front  and  to  the  right  of  the  machine.  It  will 
be  seen  that  this  view  resembles  a  photograph  with  its  lights 
and  shadows  omitted.  Perspective  views  are  much  more 
valuable  in  many  cases  for  illustrating  purposes  than  photo¬ 
graphic  reproductions,  since  they  show  clearly  parts  that  in 
a  photograph  would  be  in  the  shadow  and,  hence,  indistinctly 
perceived.  _ 


DIAGRAMMATIC  VIEWS 

22.  Diagrammatic  views  are  those  that  do  not  show 
the  exact  structure  of  an  object,  but  are  simply  used  for  indi- 


Dra/t  Change  6ear—g8/~ 

£ /£ | 

e20 

Fig.  20  L_L_ 

eating  it,  or  indicating  its  position  or  some  other  special 
feature.  It  is  difficult  to  lay  down  any  definite  definition  of 
a  diagrammatic  drawing,  since  a  great  many  types  of  these 
drawings  and  many  methods  of  indication  are  used.  In 
general,  the  purpose  of  a  diagrammatic  drawing  is  more  that 
of  indicating  the  principles  on  which  the  action  of  a  mech¬ 
anism  are  based  than  its  construction.  This  class  of  draw¬ 
ings  indicates  therefore  the  elements  of  the  mechanism 
mostly  by  means  of  conventional  symbols  instead  of  giving 
the  correct  outlines  and  proportions  of  the  parts.  It  is  also 
allowable  to  change  the  positions  of  the  parts  and  even  to 
eliminate  some  parts  altogether,  if  this  tends  to  make  the 


§91 


READING  TEXTILE  DRAWINGS 


21 


drawing  more  clear.  Fig.  20  shows  one  form.  It  is  a  rep¬ 
resentation  of  four  pairs  of  drawing  rolls  a,b,c,d,  together 
with  the  method  of  driving  them  by  gears,  and  the  sizes  of 
the  gears  used.  The  front  roll  a  receives  motion  through 
the  tight  pulley;  the  gear  e  on  this  roll  drives  the  third  roll  c 
by  means  of  the  gears  f,g,  h\  the  fourth  roll  is  driven  from 
the  third  roll  by  the  gears/,  k,  l\  the  gear  k  is  an  idler,  or 
carrier,  gear.  The  second  roll  b  is  driven  from  the  third  roll 
by  the  gears  /,  m ,  n\  the  gear  rn  in  this  case  is  an  idler,  or 
carrier,  gear.  It  will  be  seen  that  while  this  drawing  does 
not  give  an  exact  representation  either  of  the  rolls  or  of 


the  gears  used  to  connect  them,  it  does  give  an  indication  of 
the  rolls  and  of  the  gears  that  drive  them.  Fig.  21  gives 
another  illustration  of  a  diagrammatic  drawing.  In  this  case 
a  representation  is  given  of  the  method  of  driving  a  section 
of  twelve  spindles  by  means  of  one  driving  band,  to  which 
motion  is  imparted  by  the  rotating  drum  c.  This  drawing 
does  not  give  an  exact  representation  of  the  parts,  but 
simply  shows  the  method  of  passing  the  band  around  the 
drum  and  around  the  spindles  in  order  to  drive  the  twelve 
spindles  with  one  band. 


22 


READING  TEXTILE  DRAWINGS 


§91 


(a)- 


(b) 


LINES  USED  ON  DRAWINGS 
23.  In  general,  there  are  but  six  kinds  of  lines  used  on 
drawings;  they  are  shown  in  Fig.  22.  The  light,  full  line  ( a ) 
is  used  more  than  any  of  the  others  and  is  used  for  drawing 
the  outlines  of  all  parts  of  an  object  that  can  be  seen  by  the 
eye.  The  dotted  line  ( b )  consists  of  a  series  of  very  short 
dashes.  It  is  used  for  showing  the  position  and  shape  of  an 
object  or  part  of  an  object  that  is  concealed  from  the  eye  in 
the  view  shown,  on  account  of  other  parts  of  the  object  or 

_ _  other  objects  intervening. 

The  use  of  the  dotted 
line  is  shown  in  Fig.  1. 

_ The  broken-and-dotted 

lines  (c)  and  (d) ,  Fig.  22, 
'  sometimes  called  dot- 

_ and-dasli  lines,  consist 

of  a  long  dash  followed 
by  either  one  or  two  dots, 
repeated  regularly.  It  is 
very  seldom  that  both  these  lines  appear  in  the  same 
drawing,  being  used  either  to  represent  center  lines  or  indi¬ 
cate  where  a  section  has  been  taken  when  a  sectional  view 
of  an  object  is  shown  in  another  drawing.  The  broken 
line  (e)  is  used  chiefly  for  dimension  lines;  it  consists  of  a 
series  of  long  dashes.  This  line  appears  in  Figs.  28  and  29. 
The  heavy,  full  line  (/)  is  made  not  less  than  twice  as  thick 
as  ordinary  full  lines  on  a  drawing,  and  is  used  only  for 
shade  lines. 


(c)- 


(d)- 


(e)- 


(f)- 


Fig.  22 


SHADE  LINES 

24.  On  some  mechanical  drawings,  it  will  be  observed 
that  a  number  of  the  lines  are  made  very  heavy.  These 
heavy  lines  are  put  on  in  accordance  with  a  certain  almost 
universally  adopted  system;  they  show  by  their  location 
whether  the  part  looked  at  is  a  hole  in  the  object,  or  is  raised 
beyond  the  surface.  They  are  called  shade  lines.  Thus, 
in  Fig.  23,  by  means  of  the  shade  lines  a  person  can 


91 


READING  TEXTILE  DRAWINGS 


23 


determine,  without  looking  at  any  other  view  of  the  object, 

that  the  rectangles  1  and  4  represent  square  holes,  and  2 

and  3  square  bosses.  A 

Likewise,  the  heavy 

lines  CD  and  D  B  show 

that  the  object  has  a 

material  thickness. 

25.  In  the  almost 
universally  adopted 
system  of  shade  lining, 
the  light  is  assumed  to 
come  in  one  invariable 
direction,  in  such  a  man¬ 
ner  as  to  be  parallel 
with  the  plane  of  the 
paper;  to  make  an  angle 

of  45°  with  all  horizontal  and  vertical  lines  of  the  drawing, 
and  to  come  from  the  upper  left-hand  corner  of  the  draw¬ 
ing.  Each  view  of  the 
object  represented  is 
shaded  independently  of 
any  of  the  others;  and, 
when  shading,  the  ob¬ 
ject  is  always  supposed 
to  stand  so  that  the  view 
that  is  being  shaded 
will  be  a  top  view. 

Any  surface  that  can 
be  touched  by  any  one 
of  a  series  of  parallel 
straight  lines,  drawn  at 
an  angle  of  45°  with  the 
horizontal  and  vertical 
lines  of  the  drawing,  is 
called  a  light  surface; 
a  surface  that  cannot  be  touched  by  lines  having  this  angle 
is  called  a  dark  surface.  All  the  edges  produced  by  the 


Fig.  23 


24 


READING  TEXTILE  DRAWINGS 


91 


intersection  of  a  light  and  a  dark  surface  are  usually  shade- 
lined. 

26.  An  application  of  these  rules  is  shown  in  Fig.  24, 
where  a  top  view  of  a  series  of  triangular  wedges  radiating 
from  the  common  center  O  is  given.  The  top  surfaces  are, 
of  course,  light  surfaces,  but  some  of  the  perpendicular  sur¬ 
faces,  which  are  indicated  by  the  outlines  of  the  wedges,  are 
light  and  some  dark.  Observing  the  arrows,  which  here 
represent  the  direction  of  the  light,  it  is  noticed  that  they 

strike  the  edge  O  A  in 
the  wedge  R  O  A;  this 
side  will  therefore  be  a 
light  surface,  and  as 
the  top  surface  is  also 
a  light  surface,  the 
line  O  A  is  drawn  as  a 
light  line.  OR,  on  the 
contrary,  is  a  heavy 
line,  since  the  light 
cannot  strike  this 
side  without  passing 
through  the  wedge; 
hence,  this  is  a  dark 
surface,  and  its  junc¬ 
tion  OR  with  the  light 
surface  OAR  requires 
a  shade  line.  For  the 
same  reason  A R  is  also 
shaded.  The  same  reasoning  applies  to  the  lines  OB,  O  D, 
OG,  01,  OK,  and  O  M;  also  to  Q  N,  ML,  and  K  J.  C B  is 
not  shaded,  because  the  light  strikes  the  surface  of  which  C  B 
is  the  edge.  O  N  makes  an  angle  of  exactly  45°  with  a 
horizontal  line,  and  is  treated  as  if  it  were  the  edge  of  a 
light  surface;  this  is  done  in  every  case  in  which  the  line 
considered  makes  an  angle  of  45°  with  a  horizontal  line. 

27.  The  benefit  to  be  derived  from  shade  lines  is  well 
illustrated  by  Fig.  25.  Let  ( a )  be  a  front  view,  and  ( b )  a 


Fig.  25 


§91 


READING  TEXTILE  DRAWINGS 


25 


side  view  of  an  object.  Looking  only  at  the  front  view,  it  is 
impossible  to  determine  whether  the  part  A  is  a  hole,  as 
shown  by  the  side  view  given  in  Fig.  25  (b),  or  a  boss, 
as  shown  by  the  side  view  given  in  Fig.  25  (c) .  Now,  if  the 
front  view  of  the  object  be  drawn  with  heavy  shade  lines 
placed  as  in  Fig.  25  (d) ,  the  shade  lines  by  their  position 
immediately  show  that  the  part  A  is  a  hole,  without  having 
first  to  refer  to  the  side  view  shown  at  (b).  If  the  shade 
lines  are  placed  as  in  Fig.  25  (e) ,  they  show  that  the  part  A 
is  a  boss,  without  any  reference  to  the  side  view. 


BREAKS 

28.  When  a  long  and  comparatively  slender  object  is  to 
be  drawn,  it  often  happens  that,  when  drawn  to  a  sufficiently 
large  scale  to  make  it  intelligible,  it  will  extend  beyond  the 
space  available.  In  such  a  case,  part  of  the  object  is  shown 


Fig.  26 


broken  out  and  the  remaining  ends  are  close  to  each  other. 
The  fact  that  part  of  the  object  is  broken  away  for  the  sake 
of  convenience  is  indicated  by  a  so-called  break.  It  is 
always  understood  that  the  part  broken  away  and  not  shown 
is  of  the  same  transverse  dimensions  and  shape  as  the  parts 
contiguous  to  the  break.  In  some  cases,  one  end  of  the 
object  is  broken  away. 

Fig.  26  shows  a  common  method  of  drawing  a  rod  or  any 
similar  object  too  long  for  the  space  available.  It  being- 
essential  in  the  particular  case  shown  to  know  the  shape  and 
size  of  the  ends,  the  rod  is  therefore  shown  with  its  central 
part  broken  away,  as  indicated  by  the  breaks  at  a  and  b.  It 
is  customary  always  to  section  the  break  to  suit  the  material 
of  which  the  object  is  made. 

29.  Breaks  may  be  indicated  in  various  ways;  most  com¬ 
monly,  the  break  is  given  an  outline  that  will  reveal  the. 


26 


READING  TEXTILE  DRAWINGS 


§91 


shape  of  the  object.  Conventional  methods  of  indicating 
breaks  are  shown  in  Fig.  27.  Wood  is  usually  shown  broken 

in  the  manner  illustrated  at  (a);  angle 
irons,  as  at  ( b)\  T  irons,  as  at  ( c ); 
Z  bars,  as  at  (d) .  Cylindrical  objects 
are  occasionally  broken  as  shown  at  (e) , 
but  most  frequently  in  the  manner 
shown  at  (/).  Pipes  and  similar  hol¬ 
low  cylindrical  objects  may  be  broken 
as  shown  at  (g),  but  more  frequently 
the  break  is  made  as  shown  at  (/i). 
Rectangular  objects  may  be  broken  in 
the  manner  shown  at  (z);  plates  and 
objects  other  than  those  included 
between  views  (a)  and  (z),  are  often 
shown  broken  off  by  drawing  a  wavy 
freehand  line  as  in  (/)  and  (k). 


(b) 


(o) 


(d) 


(e) 


(f) 


(O) 


(h) 


d) 


(j) 


(k) 

Fig.  27 


-I 

3. 

□ 


~1 


5 

] 


SCALES 


30.  Drawings  of  an  object  may  be 
made  the  actual  size  of  the  object,  may 
be  enlarged  so  as  to  be  larger  than  the 
object  itself,  or,  as  is  generally  the  case, 
may  be  made  smaller  than  the  object 
itself.  When  a  drawing  of  an  object  is 
not  made  the  exact  size  of  the  object, 
it  is  said  to  be  drawn  to  a  scale.  For 
instance,  suppose  that  it  is  required  to 
make  a  drawing  one-quarter  the  size 
of  the  object  to  be  represented,  so  that 
3  inches  on  the  drawing  will  represent 
1  foot  on  the  object.  Then,  if  3  inches 
is  laid  off  and  divided  into  twelve  equal 
parts,  each  part,  which  is  actually  i  inch 
in  length,  will  represent  1  inch  on  the  object.  If  each  part  is 
divided  into  eight  parts,  each  part  will  represent  i  inch  on  the 
object.  A  scale  of  this  kind  is  called  a  quarter  scale,  or  a  scale 


§91 


READING  TEXTILE  DRAWINGS 


27 


of  3  inches  to  the  foot.  An  eighth  scale ,  or  a  scale  of  1  k  inches 
to  the  foot ,  would  be  constructed  in  a  similar  manner,  except 
that  lv  inches  would  be  laid  off  and  subdivided,  instead  of 
3  inches.  In  cases  where  small  objects  are  drawn  to  an 
enlarged  scale,  the  scale  might  be  18  inches  to  the  foot,  in 
which  case  the  object  would  be  drawn  li  times  its  real  size, 
or  if  the  scale  had  24  inches  to  the  foot,  the  drawing  would 
be  twice  its  actual  size. 


DIMENSION  LINES 

31.  Dimension  lines  are  used  on  drawings  to  show  the 
size,  or  dimensions,  of  the  object;  they  are  especially  neces¬ 
sary  when  objects  are  drawn  larger  or  smaller  than  they 
actually  are.  Dimension  lines  may  be  placed  directly  on  an 
object,  or  they  may  be  placed  outside  of  the 
object,  as,  for  instance,  the  dimension  line 
in  Fig.  28  marked  inches,  which  shows  * 

that  the  length  of  the  cop  there  shown  is 
62  inches.  In  the  latter  case,  extension, 
or  limiting,  lines,  as  a  and  b,  are  drawn. 

The  dimension  line  is  then  placed  between 
them,  and  arrowheads  are  placed  on  its 
ends  in  contact  with  the  limiting  lines,  as 
shown.  This  indicates  that  the  measure¬ 
ment  is  made  between  the  points  from 
which  the  limiting  lines  have  been  drawn. 

Limiting  lines  c ,  d,  e,  f  are  also  drawn  in 

Fig.  28,  and  the  dimension  lines  drawn  to 

them  show  that  the  length  of  taper  of  the 

cop  bottom  is  f  inch,  the  length  of  the  body  ^ 

of  the  cop  4i  inches,  and  the  length  of  the 

taper  at  the  nose  li  inches. 

As  a  further  illustration  of  the  use  of  FlG‘ 28 

dimension  lines,  Fig.  29  ( a )  and  ( b )  is  given,  (a)  showing  a 
plan  view  and  (b)  a  side  elevation  of  a  shuttle.  According  to 
the  dimensions  given,  the  entire  length  of  the  shuttle  is 
17  inches;  its  greatest  width  measured  across  the  top  is 
lit  inches;  the  thickness  of  its  sides  is  i  inch;  the  entire 


Fig.  28 


28 


READING  TEXTILE  DRAWINGS 


§91 


length  of  the  hollow  portion  is  9H  inches,  and  this  hollow  is 
located  4  inches  from  the  heel  of  the  shuttle.  The  length  of 
the  spindle  in  the  shuttle  is  7  inches,  and  its  diameter  varies 


from  f  inch  at  the  base  to  iV  inch  at  the  tip.  The  height  of 
the  shuttle  is  1|  inches  and  the  groove  on  the  side  is  inch  in 
width;  the  eye  of  the  shuttle  is  located  3f  inches  from  the  tip. 


CONVENTIONAL  METHODS  OF  REPRESENTING 

OBJECTS 

32.  Owing  to  the  fact  that  certain  parts  occur  with  great 
frequency  in  mechanical  drawings  and  that  considerable  time 
is  required  to  draw  them  in  the  regular  manner,  it  is  prefer¬ 
able  to  represent  them  by  certain  conventional  methods. 
Take  for  instance  screw  threads  and  tapped  holes,  as  shown 
in  Fig.  30. 

Referring  to  the  figure,  probably  the  clearest  representa¬ 
tion  of  a  screw  thread  is  shown  in  Fig.  30  (a)  and  (e) . 
Fig.  30  (a)  is  a  full  view  of  a  screw  (a  bolt  in  this  instance); 
Fig.  30  («?)  is  a  sectional  view  of  a  tapped  hole.  Fig.  30  ( b ) 
and  (/)  shows  a  representation  of  a  screw  thread  and  tapped 
hole  that  may  be  drawn  more  easily  and  still  give  a  good 
representation  of  the  thread.  Fig.  30  ( c )  and  (g)  shows  a 
conventional  method  of  representing  a  screw  thread  and 
tapped  hole  that  is  very  largely  used.  While  it  does  not  by 
any  means  show  the  exact  appearance  of  the  thread,  it  readily 


§91 


READING  TEXTILE  DRAWINGS 


29 


conveys  to  the  mind  what  the 
Fig.  30  ( d )  and  (k)  illustrates 


drawing  is  intended  to  show, 
another  and  still  more  simple 


(c)  (d) 


method  of  representing  a  screw  thread  and  tapped  hole, 
but  this  method  is  so  crude  that  it  is  little  used  except 
in  drawings  where  the  appearance  must  be  sacrificed  to 


(a) 


(t» 


(c) 


(d) 


Fir,.  31 


rapidity  of  execution  on  the  part  of  the  draftsman.  It  will 
be  noticed  that  in  each  case  the  slant  of  the  thread  in  the  hole 
is  in  the  opposite  direction  to  that  on  the  screw. 


30 


READING  TEXTILE  DRAWINGS 


§91 


33.  When  the  screw  thread  is  hidden  by  part  of  the 
object,  and  it  is  deemed  necessary  to  show  it  in  dotted  lines, 
it  is  usually  drawn  in  one  of  the  four  ways  illustrated  in 
Fig-.  31.  Of  these,  the  method  shown  in  Fig.  31  (a)  is  prob¬ 
ably  the  clearest;  that  shown  in  Fig.  31  (b)  is  fairly  good;  and 
the  one  shown  in  Fig.  31  ( c )  may  be  drawn  quickly  but  is  not 
a  good  representation.  The  method  illustrated  in  Fig.  31  ( d ) 
is  practically  no  representation  of  a  screw  thread  at  all. 

34.  All  the  threads  shown  in  Figs.  30  and  31  are  riglit- 
handed;  that  is,  when  looking  along  the  axis  of  the  screw, 
the  thread  advances  in  the  direction  in  which  the  hands  of  a 

watch  move.  When  the  thread  advances 
in  an  opposite  direction,  the  thread  is 
left-handed;  it  is  then  drawn  with 
the  thread  slanting  the  other  way  from 
that  shown  in  Figs.  30  and  31.  Thus, 
in  Fig.  32  a  left-handed  screw  thread  is 
shown  in  full,  and  a  left-handed  tapped 
hole  in  section  beneath  it.  It  will  be 
noticed  that  the  thread  in  the  hole 
slants  the  opposite  way  to  that  on  the 
screw. 

In  case  of  doubt,  a  right-handed 
thread  can  always  be  told  from  a  left- 
handed  thread  by  holding  the  screw, 
or  the  drawing  of  it,  so  that  the  axis 
is  vertical.  Then,  if  the  thread  slants 
upwards  to  the  right  in  case  of  a  screw  shown  in  full,  or 
upwards  to  the  left  in  case  of  a  tapped  hole  in  section,  the 
thread  is  right-handed;  if  otherwise,  it  is  left-handed. 

35.  Springs  also  are  frequently  represented  by  con¬ 
ventional  methods;  for  instance,  Fig.  33  (a)  shows  a  method 
of  representing  a  spring  that  practically  gives  its  exact 
appearance,  but  as  this  method  of  showing  springs  is  difficult 
to  draw,  conventional  methods,  as  shown  at  Fig.  33  (b)  and 
(c),  are  frequently  used.  In  addition,  chains  are  frequently 
represented  on  a  drawing  in  some  conventional  manner. 


§91 


READING  TEXTILE  DRAWINGS 


31 


Fig.  34  (a)  shows  a  method  of  representing  a  chain  that  gives 
its  exact  appearance;  Fig.  34  (b)  shows  a  method  of  indi¬ 
cating  a  small  chain  that  may  be  much  more  readily  drawn 


Fig.  33 


0  !  < 

r) 

'o' 

v )  i,i 

n  f  ( 

OJ 

& 

P  i  , 

n  1  1 

o) 

'O' 

in  |  « 

O) 

& 

a  < 

6) 

'o' 

I  |  , 

o ) 

& 

OJ 

'o' 

U  A  ( 

o) 

V& 

(a)  (b)  (c)  (d) 


Fig.  34 


by  the  draftsman.  Fig.  34  (c)  shows  the  exact  appearance 
of  a  sprocket  chain,  while  Fig.  34  (d)  is  a  conventional  repre¬ 
sentation  that  is  frequently  used.  Conventional  representa¬ 
tions  of  ropes  are  often  used.  Fig.  35  (a)  shows  the  exact 


(a) 


(b) 


Fig.  35 

appearance  of  a  rope,  while  Fig.  35  {b)  is  a  conventional  repre¬ 
sentation  that  is  frequently  used.  The  object  of  using  all 
these  conventional  methods  of  representing  various  objects  is, 
of  course,  to  save  the  draftsman’s  time  in  making  drawings. 


Fig.  36 


§91 


READING  TEXTILE  DRAWINGS 


33 


MOVING  PARTS  SHOWN  IN  TWO  OR  MORE 

POSITIONS 

36.  Dotted  lines,  as  previously  explained,  are  used  to 
show  those  parts  of  an  object  that  are  concealed  from  the 
eye  in  the  view  of  the  object  given,  by  other  parts  interve¬ 
ning  between  them  and  the  eye.  In  addition,  they  are  used  to 
indicate  parts  of  mechanism  when  it  is  desired  to  show  them 
in  more  than  one  position.  This  use  of  dotted  lines  is 
merely  for  the  purpose  of  indicating  on  the  drawing  the 
extent  of  the  movement  of  certain  parts  or  the  positions  that 
they  assume  under  certain  conditions;  as  an  illustration  of 
this  use  of  dotted  lines  Fig.  36  is  given.  This  is  a  view  of  the 
shedding  mechanism  of  a  cam-loom.  The  harness  m  and  the 
jack  l  are  shown  in  one  position  in  full  lines,  while  in  order 
to  show  the  extent  of  their  movement,  they  are  shown  in 
dotted  lines  in  the  position  that  they  assume  when  the 
harness  is  lowered,  at  which  time  the  cam-bowl  /2  on  the 
jack  is  in  its  position  nearest  to  the  cam-shaft  kx:  The  full 
lines  represent  the  position  of  the  harness  and  the  jack  when 
the  harness  is  at  its  highest  position,  while  the  dotted  lines 
show  their  position  when  the  harness  is  in  its  lowest  position. 


REFERENCE  TETTERS 

37.  Reference  letters  are  placed  on  the  various  parts 
shown  in  a  drawing  so  that  these  parts  can  be  definitely 
referred  to  in  the  accompanying  text.  There  are,  in  general, 
three  ways  of  placing  reference  letters  on  a  drawing. 
The  letter  may  be  placed  directly  on  the  object  to  which 
reference  is  to  be  made,  as  for  instance  in  the  case  of  the 
letter  h,  Fig.  19,  where  the  reference  letter  is  placed  directly 
on  the  take-up  roll  of  the  loom.  A  reference  letter  may  be 
placed  to  one  side  of  the  object  to  be  referred  to  and  a 
line  drawn  to  the  object,  as  for  instance  in  the  case  of  the 
letter  k2,  Fig.  19,  where  the  reference  letter  refers  to  the  reed 
cap  of  the  loom.  Or,  where  confusion  will  not  result,  the 
letters  are  placed  beside  the  object  to  which  reference  is 


READING  TEXTILE  DRAWINGS 


34 


§91 


to  be  made  and  the  line  omitted,  as  in  the  case  of  the 
letter  v  at  the  shipper  handle  in  Fig.  19. 

When  a  machine  has  a  series  of  parts  that  cooperate  in 
such  a  manner  as  to  constitute  a  definite  mechanism,  more  or 
less  complete  in  itself,  each  part  of  such  a  combination  is 
referred  to  by  the  same  letter,  distinction  being  made 
between  the  several  parts  by  giving  the  letter  different  sub- 
scripts.  For  instance,  suppose  that  a  certain  mechanism 
contains  a  train  of  three  gears;  these  gears  would  be  let¬ 
tered  a,a1}aa,  or  if  more  gears  were  contained  in  the  same 
mechanism,  the  subscripts  would  run  still  higher  until  each 
part  referred  to  in  the  text  was  lettered. 

Where  several  drawings  are  used  to  illustrate  the  same 
machine  or  mechanism,  the  same  part,  if  shown  in  more  than 
one  of  these  drawings,  always  has  the  same  reference  letter, 
except  in  special  cases.  Thus,  a  gear  that  is  marked  a  in 
one  drawing  will  be  marked  a  in  every  other  drawing  in  which 
it  may  appear. 

When  reference  letters  are  enclosed  in  parentheses 
thus  (a),  (b) ,  etc.,  the  reference  may  be  to  a  separate  view 
of  some  particular  element  of  a  mechanism,  but  is  mostly  to 
different  views  of  the  entire  mechanism.  For  instance,  if 
reference  is  made  to  Fig.  29  (a),  it  is  desired  to  call  atten¬ 
tion  to  that  view  of  the  object  marked  (a)  in  Fig.  29. 


A  SERIES  OF  QUESTIONS 


Relating  to  the  Subjects 
Treated  of  in  This  Volume. 


It  will  be  noticed  that  the  questions  contained  in  the  fol¬ 
lowing  pages  are  divided  into  sections  corresponding  to  the 
sections  of  the  text  of  the  preceding  pages,  so  that  each 
section  has  a  headline  that  is  the  same  as  the  headline  of 
the  section  to  which  the  questions  refer.  No  attempt  should 
be  made  to  answer  any  of  the  questions  until  the  corre¬ 
sponding  part  of  the  text  has  been  carefully  studied. 


ARITHMETIC 

(PART  1) 


EXAMINATION  QUESTIONS 

(1)  In  a  certain  year,  Great  Britain  used  4,141,000  bales 

of  cotton,  while  the  United  States  used  2,812,000  bales;  how 
many  bales  of  cotton  were  used  by  both  of  these  countries  in 
that  year?  Ans.  6,953,000  bales 

(2)  In  12  months,  Russia  used  778,000  bales  of  cotton, 
France  650,000,  Germany  1,170,000,  Austria  530,000,  Switzer¬ 
land  150,000,  Sweden  80,000,  Holland  65,000,  Belgium 
145,000,  Spain,  Portugal,  and  Greece  together  360,000,  and 

Italy  390,000;  what  was  the  total  number  of  bales  used  by 

✓ 

these  countries?  Ans.  4,318,000  bales 


(3)  In  a  certain  year,  the  number  of  bales  of  cotton 
raised  by  North  Carolina  was  336,261,  while  South  Carolina 
produced  747,190;  how  many  more  bales  were  produced  by 
South  Carolina  than  by  North  Carolina?  Ans.  410,929  bales 

(4)  In  a  certain  year,  Georgia  produced  1,191,864  bales 

of  cotton  and  Mississippi  1,154,725;  what  was  the  total 
number  of  bales  produced  by  Georgia  and  Mississippi  in  this 
year?  Ans.  2,346,589  bales 


(5)  Solve  the  following  problems: 

(a)  436  +  14  +  89  +  1,075;  (b)  100,173  +  19  +  11  +  8  +  53. 

A ns  J(«)  1-614 
AnsT(<M  100,264 


n 


2 


ARITHMETIC 


§1 


(6)  In  12  months,  Kentucky  produced  873  bales  of  cotton; 
if  the  average  weight  of  a  bale  for  the  year  was  480  pounds, 
what  was  the  number  of  pounds  produced?  Ans.  419,040  lb. 

(7)  In  a  certain  year,  Virginia  produced  5,375  bales  of 
cotton;  if  the  average  weight  of  a  bale  was  485  pounds,  what 
was  the  number  of  pounds  produced?  Ans.  2,606,875  lb. 


(8)  According  to  the  statistics  for  a  certain  year,  Massa¬ 

chusetts  had  187  cotton  mills  that  were  operating  5,824,489 
spindles;  what  was  the  average  number  of  spindles  to  each 
mill?  Ans.  31,147  spindles 

(9)  The  number  of  looms  run  by  187  mills  was  133,144; 
what  was  the  average  number  of  looms  to  each  mill? 

Ans.  712  looms 

(10)  The  number  of  bales  of  cotton  consumed  by  187 

mills  was  772,497;  what  was  the  average  number  of  bales 
used  by  each  mill?  Ans.  4,131  bales 


(11)  The  number  of  operatives  in  the  Massachusetts  mills 
was  76,213  and  the  amount  of  money  paid  to  them  was 
26,217,272  dollars;  what  was  the  average  amount  of  money 
•received  by  each  person?  Ans.  344  dollars 


(12) 
( a ) 


Solve  the  following  problems: 
278,467  -  28,903;  ( b )  478,096  - 


203,457. 

f(fl)  249,564 
'!(£)  274,639 


Ans. 


(13)  In  a  certain  year,  the  United  States  produced  309,- 
474,856  pounds  of  wool,  and  in  the  next  year  294,296,726 
pounds;  how  many  more  pounds  were  produced  in  the  first 
year  than  in  the  second?  Ans.  15,178,130  lb. 


(14)  In  a  certain  year,  there  were  in  Maine  247,168  sheep 
and  the  amount  of  wool  produced  was  1,483,008  pounds; 
what  was  the  average  weight  of  wool  taken  from  each  sheep? 

Ans.  6  lb. 


(15)  Solve  the  following  problem:  337,809  X  448. 

Ans.  151,338,432 


§1 


ARITHMETIC 


3 


(16)  If  17,315,097  pounds  of  wool,  after  being  scoured, 

weighed  5,540,831  pounds,  what  weight  was  lost  in  the 
scouring?  Ans.  11,774,266 

(17)  Solve  the  following  problem:  231,539  -r-  97. 

Ans.  2,387 

(18)  In  a  certain  year,  North  Carolina  produced  1,117,485 

pounds  of  wool  from  223,497  sheep;  what  was  the  average 
number  of  pounds  produced  by  each  sheep?  Ans.  5  lb. 

(19)  Solve  the  following  expressions  by  cancelation: 

,  .  22  X  12  X  5x9  ?  ,  *  72  X  12  X  81  X  26  _  ? 

U  6X  3X11X4  '  36  x  54  x  13  x  24 

Ans  I  ^  15 
Ans-\(£)  3 

(20)  How  many  spindles  are  there  in  a  woolen  mill  that 

has  8  mules  with  280  spindles  each  and  14  mules  with  320 
spindles  each?  Ans.  6,720  spindles 


ARITHMETIC 


(PART  2) 


EXAMINATION  QUESTIONS 

(1)  What  will  be  the  cost  of  28  ring-spinning  frames  of 
208  spindles  each  at  $3.15  per  spindle?  Ans.  $18,345.60 

(2)  How  many  pounds  of  cotton  are  there  in  50  yards  of 

union  cloth  weighing  1.3  pounds  per  yard,  if  f  of  the  cloth  is 
cotton,  the  rest  being  wool?  Ans.  26  lb. 

(3)  What  is  the  wool  clip  from  2,317,636  sheep  if  the 
fleeces  average  6.75  pounds  in  weight?  Ans.  15,644,043  lb. 

(4)  Multiply  4.02  by  .565  and  divide  the  product  by  .75. 

Ans.  3.028 

(5)  If  the  average  weight  of  Ohio  fleeces  is  5.75  pounds, 

how  many  sheep  will  it  take  to  grow  a  bale  of  wool  weighing 
552  pounds?  Ans.  96  sheep 

(6)  What  is  the  value  of  41,883,065  sheep  if  the  average 

value  of  a  single  sheep  is  $2.93?  Ans.  $122,717,380.45 

(7)  There  are  437.5  grains  in  1  ounce;  how  many  ounces 
will  20  yards  of  sliver,  weighing  220  grains  per  yard,  weigh? 

Ans.  10.057  oz. 

(8)  If  76,213  people  engaged  in  cotton  manufacturing 

earn  for  a  certain  year  $26,230,667.48,  what  is  the  average 
amount  earned  by  each  person?  Ans.  $344,175 

(9)  If  middling  uplands  cotton  is  worth  8f  cents  per 
pound,  what  is  the  cost  of  a  bale  weighing  496  pounds? 

Ans.  $41.54 


2 


ARITHMETIC 


§2 


(10)  A  mill  was  sold  at  auction  for  $110,250,  which  was 
I  of  its  original  cost;  what  was  the  cost?  Ans.  $275,625 

(11)  What  is  the  production  per  day  of  a  weave  room 

containing  978  looms,  the  average  production  of  each  loom 
being  34i  yards?  Ans.  33,4962  yd. 

(12)  The  cotton  crop  of  the  United  States  during  a 

certain  year  was  10,533,000  bales,  of  which  f  was  exported; 
how  many  bales  were  exported?  Ans.  7,022,000  bales 

(13)  If  a  cotton  card  produces  24  pounds  of  carded  cot¬ 
ton  in  1  hour,  what  is  the  production  in  10i  hours? 

Ans.  252  lb. 

(14)  A  certain  grade  of  wool  loses  i\  of  its  weight  in 

scouring;  what  weight  would  be  lost  in  scouring  4,780  pounds 
of  this  wool?  Ans.  1,434  lb. 

(15)  What  is  the  cost  of  150  bales  of  cotton  averaging 
486  pounds  per  bale  at  101  cents  per  pound?  Ans.  $7,581.60 

(16)  At  a  certain  time,  the  price  of  XX  Ohio  wool  was 
28i  cents  per  pound  in  New  York  and  19^  cents  per  pound 
in  London;  how  much  more  money  would  American  wool 
bring  per  1,000  pounds  in  the  home  market?  Ans.  $90 

( 17 )  What  is  the  cost  of  5,850  pounds  of  wool  at  222  cents 

per  pound?  Ans.  $1,316.25 

(18)  What  is  the  value  of  a  cut  of  cloth  containing 

47i  yards  at  $1  per-  yard?  Ans.  $17$| 

(19)  What  part  of  an  acre  is  tt  of  of  an  acre? 

Ans. 

(20)  A  certain  corporation  owns  three  mill  properties  as 

follows:  Mill  No.  1  is  valued  at  i  the  worth  of  mill  No.  2; 
mill  No.  2  is  valued  at  I2  times  the  worth  of  mill  No.  3; 
mill  No.  3  is  worth  $85,000;  what  are  the  values  of  mills 
Nos.  1  and  2?  »  f  $106,250,  value  of  mill  No.  1 

ns*  1  $127,500,  value  of  mill  No.  2 


ARITHMETIC 


(PART  3) 


EXAMINATION  QUESTIONS 

(1)  If  2s  pounds  of  cotton  is  lost,  during  transportation, 
from  a  bale  of  500  pounds,  what  is  the  rate  per  cent,  of  loss? 

Ans.  2% 

(2)  In  a  certain  year  the  cotton  crop  of  the  United  States 
was  approximately  9,400,000  bales;  of  this  the  United  States 
used  40%;  how  many  bales  were  used  in  this  country? 

Ans.  3,760,000  bales 

(3)  In  a  certain  year  7,532,000  bales  of  cotton  were  pro¬ 

duced  by  the  United  States;  of  this  amount  the  Northern 
mills  used  1,580,000  bales  and  the  Southern  mills  711,000 
bales;  what  per  cent,  of  the  entire  crop  did  the  mills  of  the 
United  States  take?  Ans.  30.41% 

(4)  In  a  certain  year,  North  Carolina  had  1,311,708  acres 

of  land  devoted  to  cotton  raising,  which  was  93%  of  the 
number  of  acres  devoted  to  cotton  during  the  season  of 
1897-98;  how  many  acres  of  land  were  used  for  this  purpose 
in  the  season  of  1897-98?  Ans.  1,410,439  acres,  nearly 

(5)  800  is  66f%  of  what  number?  Ans.  1,200 

(6)  If  a  mill  is  sold  for  $450,000,  and  this  is  75%  of  its 
original  cost,  what  was  the  cost  of  building  the  mill? 

Ans.  $600,000 

(7)  If  a  mill  has  capital  stock  to  the  value  of  $600,000, 

and  for  a  certain  year  declares  quarterly  dividends  of  12%, 
what  will  be  the  amount  of  money  paid  in  dividends  for  that 
year?  Ans.  $36,000 


83 


2 


ARITHMETIC 


§3 


(8)  Solve  the  following:  ( a )  120  :  x  —  90  :  500;  ( b )  66| 

:  100  =  x  :  90.  A  J  (a)  666f 

Ans,l(^)  60 

(9)  If  800  looms  produce  3,200  cuts  of  cloth  in  1  week, 
how  many  looms  will  be  required  if  it  is  desired  to  increase 
the  production  to  4,800  cuts  per  week?  Ans.  1,200  looms 

(10)  Solve  the  following:  (a)  100  :  25  —  40  :  x;  ( b )  x 

(a)  10 

(b)  30 

(11)  If  $25  per  year  is  paid  for  interest  on  $1,000,  what 

is  the  rate  per  cent.?  Ans.  2i% 

(12)  If  $50  per  year  is  paid  for  interest  on  a  certain  sum 
of  money,  the  rate  being  6%,  what  is  the  principal? 

Ans.  $833. 33i 

(13)  What  is  3i%  of  52  yards?  Ans.  1.69  yd. 

(14)  What  per  cent,  of  5  is  .2?  Ans.  4% 

(15)  If  17  looms  produce  680  yards  of  cloth  per  day,  how 

many  yards  will  283  looms  produce  in  the  same  length 
of  time?  Ans.  11,320  yd. 


90  =  20  :  60. 


Ans 


•{ 


ARITHMETIC 

(PART  4) 


EXAMINATION  QUESTIONS 

(1)  From  the  cube  of  3  subtract  the  square  root  of  729. 

Ans.  0 

(2)  Find  the  value  of  84.  Ans.  4,096 

(3)  The  standard  number  of  turns  of  twist  for  cotton  warp 

yarn  is  equal  to  4.75  multiplied  by  the  square  root  of  the 
counts;  how  many  turns  per  inch  will  be  placed  in  36s  counts 
if  standard  twist  is  inserted?  Ans.  28.5 


(4)  What  is  the  fourth  power  of  i?  Ans.  tAA 

(5)  Find  the  value  of  Vl7,688.  Ans.  132.996 

(6)  What  is  the  square  of  2,002?  Ans.  4,008,004 

(7)  If  the  number  of  turns  of  twist  that  is  placed  in 
roving  is  equal  to  1.2  multiplied  by  the  square  root  of 
the  hank  roving,  what  will  be  the  twist  of  5-hank  roving? 

Ans.  2.683 


(8)  Extract  the  square  root  of  .3364.  Ans.  .58 

(9)  Find  the  value  of  .04062.  Ans.  .00164836 

(10)  The  standard  number  of  turns  of  twist  for  cotton 

filling  yarn  is  equal  to  3.25  multiplied  by  the  square  root 
of  the  counts;  what  twist  should  be  inserted  in  50s  filling, 
standard  twist?  Ans.  22.98 

(11)  Extract  the  square  root  of  625.  Ans.  25 

(12)  What  is  the  square  of  .001  Ans.  .000001 


§4 


2 


ARITHMETIC 


§4 


(13)  From  the  cube  of  30  subtract  the  square  of  30. 

Ans.  26,100 

(14)  Divide  the  cube  of  7  by  the  square  root  of  49. 

Ans.  49 

(15)  What  is  the  square  root  of  676?  Ans.  26 


ARITHMETIC 

(PART  5) 


EXAMINATION  QUESTIONS 

(1)  If  it  takes  72  picks  to  make  1  inch  of  cloth  and  the 

loom  puts  in  80  picks  per  minute,  how  long-  will  it  take 
to  weave  1  yard  of  cloth?  Ans.  32  min.  24  sec. 

(2)  If  10  lb.  8  oz.  of  worsted  yarn  costs  $11.16,  what  is 

the  price  per  pound?  Ans.  $1,062 

(3)  If  a  yard  of  a  certain  cloth  weighs  20  ounces  (avoir¬ 
dupois),  how  many  pounds  will  a  cut  of  54  yards  weigh? 

Ans.  67 2  lb. 

(4)  If  a  card  produces  24  pounds  in  87  minutes,  how 
many  pounds  will  be  produced  in  a  day  of  10  hours? 

Ans.  165.51  lb. 

(5)  If  a  picker  makes  100  laps  in  13  hours  and  20  min¬ 
utes,  how  many  laps  will  it  make  in  60  hours? 

Ans.  450  laps 

(6)  Multiply  7  gal.  3  qt.  1  pt.  2  gi.  by  5. 

Ans.  1  bbl.  8  gal.  1  pt.  2  gi. 

(7)  Reduce  24,789  ounces  (avoirdupois)  to  higher  denom¬ 
inations.  Ans.  15  cwt.  49  lb.  5  oz. 

(8)  If  10  cents  will  buy  li  ounces  of  card-clothing  tacks, 
what  quantity  can  be  obtained  for  $1.25?  Ans.  1  lb.  2f  oz. 

(9)  A  loom  weaves  3  yd.  6  in.  of  a  certain  cloth  in 

50  minutes;  how  many  yards  will  be  produced  in  a  week  of 
58  hours?  Ans.  220.4  yd. 

§5 


2 


MENSURATION 


§6 


(9)  If  a  square  yard  of  cloth  weighs  20  ounces  (avoirdu¬ 

pois),  how  many  pounds  will  a  cut  of  50  yards  weigh,  the 
cloth  being  54  inches  wide?  Ans.  93.75  lb. 

(10)  How  many  cubic  feet  will  a  cylindrical  tank  4  feet 
in  diameter  and  5  feet  deep  contain?  Ans.  62.832  cu.  ft. 

(11)  How  many  bags  of  wool  weighing  400  pounds  each 

can  be  stored  in  a  room  200  feet  long  and  100  feet  wide, 
without  loading  the  floor  more  than  250  pounds  to  the  square 
foot?  .  Ans.  12,500  bags 

(12)  How  many  gallons  of  oil  are  there  in  a  cylindrical 

tank  22  inches  in  diameter  if  the  depth  of  the  oil  in  the  tank 
is  15i  inches?  Ans.  25.5068  gal. 

(13)  If  a  cubic  inch  of  lead  weighs  .41  pound,  what  is 
the  weight  of  a  ball  of  lead  7  inches  in  diameter? 

Ans.  73.634  lb. 

(14)  A  sphere  is  16  inches  in  diameter;  what  is  its  sur¬ 
face  area?  Ans.  804.250  sq.  in. 

(15)  What  is  the  average  weight  per  square  foot  on  a 

floor  carrying  150  cards,  each  weighing  5,000  pounds,  if  the 
room  is  125  feet  wide  and  300  feet  long?  Ans.  20  lb. 


MECHANICAL  DEFINI¬ 
TIONS 


EXAMINATION  QUESTIONS 

(1)  ( a )  What  is  a  line  shaft?  ( b )  What  is  a  counter¬ 
shaft? 

(2)  Describe  a  box,  or  muff,  coupling. 

(3)  What  advantage  has  a  friction  coupling  over  an 
ordinary  clutch  coupling? 

(4)  Describe  an  ordinary  shaft  hanger. 

(5)  Define  the  following:  (a)  wall  box;  (b)  pillow- 
block;  ( c )  floor  stand;  (d)  wall  bracket;  ( e )  post  hanger; 
(/)  bearing. 

(6)  What  can  be  said  about  the  distance  apart  that 
hangers  should  be  located? 

(7)  What  is  the  object  of  crowning  a  pulley? 

(8)  In  arranging  the  pulleys  for  a  quarter-turn,  how  can 
it  be  determined  when  they  are  in  the  proper  position? 

(9)  Define  the  following:  (a)  split  pulley;  (b)  flange 
pulley;  ( c )  sheave  pulley;  (d)  guide  pulley;  ( e )  binder 
pulley;  (/)  drum. 

(10)  What  are  step  pulleys  and  for  what  reason  are 
they  used  ? 

( 11 )  What  kind  of  belting  is  most  suitable  for  use  in  places 
where  the  atmosphere  is  constantly  full  of  steam  and  moisture? 


2  MECHANICAL  DEFINITIONS  §7 

(12)  ( a )  What  is  a  double  belt?  (b)  How  much  stronger 
is  a  double  belt  than  a  single  one? 

« 

(13)  («)  What  part  of  a  gear  tooth  is  called:  (a)  the 
addendum?  (b)  the  root  ? 

(14)  (<?)  Would  a  better  drive  be  obtained  with  30-inch 
pulleys  between  two  parallel  shafts  in  the  same  horizontal 
plane  if  the  shafts  were  7  feet  apart  than  when  25  feet  apart, 
all  conditions  being  normal?  (b)  Which  part  of  the  belt,  in 
case  it  is  an  open  belt,  would  you  prefer  to  have  for  the 
tight  side? 

(15)  Name  some  of  the  advantages  of  rope  transmission. 

(16)  What  is:  (a)  a  spur  gear?  (b)  a  bevel  gear?  (r)  a 
ratchet  gear?  (d)  a  sprocket  gear?  ( e )  a  pinion  gear?  (/)  an 
annular  gear? 

(17)  ( a )  How  are  elliptic  and  eccentric  gears  marked  for 
setting?  (£)  Why  is  this  necessary? 

(18)  (a)  What  is  the  pitch  circle  of  a  gear?  {b)  What 
is  the  circular  pitch  of  a  gear?  (c)  What  is  the  diametral 
pitch  of  a  gear? 

(19)  (a)  What  is  a  positive-motion  cam?  ( b )  What  is  a 
non-positive  cam?  (r)  What  is  meant  by  the  terms  heel 
and  toe  of  a  cam? 

(20)  State  the  relative  positions  of  the  power,  weight, 
and  fulcrum  in  levers  of  the  first,  second,  and  third  classes. 


MECHANICAL  CALCU¬ 
LATIONS 


EXAMINATION  QUESTIONS 

(1)  A  36-inch  driving  pulley  is  fastened  to  a  shaft  making 
275  revolutions  per  minute  and  drives  a  16-inch  pulley  on  a 
machine.  What  is  the  speed  of  the  machine? 

Ans.  618.75  rev.  per  min. 

(2)  The  shaft  of  a  new  machine  should  be  driven  560 

revolutions  per  minute  and  is  equipped  with  16-inch  tight 
and  loose  pulleys.  The  driving  shaft  from  which  the 
machine  is  to  be  driven  makes  320  revolutions  per  minute. 
How  large  a  driving  pulley  must  be  ordered  to  drive  the 
machine  at  the  required  speed?  Ans.  28-in.  pulley 

(3)  A  driving  shaft  carries  a  24-inch  pulley  and  makes 

420  revolutions  per  minute;  the  driven  pulley  makes  600  revo¬ 
lutions  per  minute;  what  is  its  diameter?  Ans.  16.8  in. 

(4)  A  20-inch  pulley,  on  a  main  shaft  making  280  revo¬ 

lutions  per  minute,  drives  a  24-inch  pulley  on  a  countershaft; 
attached  to  the  countershaft  there  is  also  a  9-inch  pulley 
driving  a  12-inch  pulley  on  the  shaft  of  a  machine;  what  is 
the  speed  of  the  machine?  Ans.  175  rev.  per  min. 

(5)  A  certain  machine  must  make  450  revolutions  per 
minute;  the  10-inch  driven  pulley  must  be  driven  from  a 
15-inch  pulley  on  a  countershaft  that  must  be  driven  from 
a  30-inch  pulley  on  a  main  shaft  making  350  revolutions  per 


2 


MECHANICAL  CALCULATIONS 


§8 


minute;  how  large  a  driven  pulley  must  be  ordered  for  the 
countershaft  in  order  to  give  the  machine  the  proper  speed? 

Ans.  35-in.  pulley 

(6)  What  size  driving  pulley  should  be  purchased  for  a 
shaft  making  480  revolutions  per  minute  to  drive  a  machine 
with  a  12-inch  driven  pulley  360  revolutions  per  minute? 

Ans.  9-in.  pulley 

(7)  An  80-tooth  gear  making  30  revolutions  per  minute 

imparts  150  revolutions  per  minute  to  a  driven  shaft;  what 
is  the  size  of  the  driven  gear?  Ans.  16-tooth  gear 

(8)  Find  the  speed  of  the  driven  shaft  in  the  following 

train:  The  first  driver  has  70  teeth  and  makes  21  revolu¬ 
tions  per  minute  and  drives  a  40-tooth  gear  on  a  stud;  com¬ 
pounded  with  this  gear  there  is  a  42-tooth  gear  driving  a 
56-tooth  gear  on  a  shaft;  fast  to  the  same  shaft  there  is 
a  56-tooth  gear  driving  a  28-tooth  pinion  gear  on  the  driven 
shaft.  Ans.  55.125  rev.  per  min. 

(9)  How  long  an  open  belt  is  required  to  connect  two 

24-inch  pulleys,  the  distance  between  the  centers  of  the  shafts 
being  17i  feet?  Ans.  41.283  ft. 

(10)  Two  36-inch  pulleys  are  to  be  connected  by  a  single 
belt  and  make  240  revolutions  per  minute;  if  25  horsepower 
is  to  be  transmitted,  how  wide  should  the  belt  be? 

Ans.  10  in.  (practically) 

(11)  A  cylinder  3  feet  in  diameter  makes  450  revolu¬ 

tions  per  minute;  what  is  the  surface  velocity,  in  feet  per 
minute?  Ans.  4,241.16  ft.  per  min. 

(12)  In  the  following  train,  what  is  the  size  of  the  first 
driving  gear  if  it  makes  48  revolutions  per  minute  and 
drives  a  24-tooth  gear  compounded  with  a  32-tooth  gear  that 
drives  a  16-tooth  gear  96  revolutions  per  minute? 

Ans.  24-tooth  gear 

(13)  The  constant  factor  of  a  train  of  gears  is  8.75;  what 
speed  of  the  driven  shaft  will  a  20-tooth  change  gear  give? 

Ans.  175  rev.  per  min. 


8 


MECHANICAL  CALCULATIONS 


3 


(14)  The  first  driver  of  a  train  of  gears  has  80  teeth  and 

makes  48  revolutions  per  minute;  the  second  driving  gear  of 
the  train  contains  24  teeth;  the  change  gear  is  also  a  driver; 
the  driven  gears  of  the  train  contain  40,  12,  and  60  teeth, 
respectively:  (a)  Find  a  constant  factor  for  the  train. 
(b)  What  size  change  gear  will  give  128  revolutions  per 
minute  of  the  driven  shaft?  »  ,  1  (a)  3.2,  constant  factor 

ns'l(£)  40-tooth  gear 

(15)  Find  the  constant  dividend  of  a  train  of  gears  com¬ 
posed  of  the  following:  three  drivers,  the  first  having  66 
teeth  and  making  90  revolutions  per  minute,  and  the  others 
having  36  and  30  teeth,  respectively;  and  three  driven  gears, 
the  first  two  having  33  and  27  teeth,  respectively,  and  the 
third  being  the  change  gear.  Ans.  7,200,  constant  dividend 

(16)  Two  48-inch  pulleys  are  to  be  connected  with  a 
double  belt  and  make  186  revolutions;  how  wide  should  the 
belt  be  to  transmit  30  horsepower?  Ans.  8  in.  (practically) 

(17)  The  constant  dividend  of  a  train  of  gears  is  2,700; 

what  speed  will  a  30-tooth  change  gear  impart  to  the  last 
driven  shaft  of  the  train?  Ans.  90  rev.  per  min. 

(18)  A  single-threaded  worm  makes  252  revolutions  per 
minute  and  drives  a  63-tooth  worm-gear;  how  many  revolu¬ 
tions  per  minute  does  the  worm-gear  make? 

Ans.  4  rev.  per  min. 

(19)  A  mangle  gear  with  85  teeth  is  driven  by  a  12-tooth 

pinion  gear  making  120  revolutions  per  minute:  (a)  How 
many  revolutions  does  the  mangle  gear  make?  (b)  How 
many  in  each  direction?  A  f  (a)  16  revolutions 

ns'l(^)  8  revolutions 

(20)  A  certain  lever  has  a  power  arm  of  60  inches  and  a 

weight  arm  of  3  inches;  what  weight  will  be  lifted,  or  pres¬ 
sure  exerted,  at  the  end  of  the  weight  arm  by  a  force  of 
40  pounds  at  the  end  of  the  power  arm?  Ans.  800  lb. 


YARN  CALCULATIONS, 

COTTON 


EXAMINATION  QUESTIONS 

(1)  What  counts  of  yarn  should  be  twisted  with  a  24s  to 
make  a  ply  yarn  equal  in  weight  to  an  8s  single?  Ans.  12s 

(2)  An  80s,  40s,  and  30s  are  twisted  together;  what  are 

the  counts  of  the  ply  yarn?  Ans.  14.117s 

(3)  How  many  yards  are  there  in  2  pounds  of  50s  cotton 

yarn?  Ans.  84,000  yd. 

(4)  Explain  what  is  meant  by  the  word  counts. 

(5)  State  what  is  meant  by  ply  yarns. 

(6)  If  368,000  yards  of  cotton  yarn  weighs  16  pounds, 

what  are  the  counts?  Ans.  27.380s 

(7)  What  counts  of  single  yarn  are  twisted  to  make  a 
2-ply  38s? 

(8)  What  is  the  weight  of  360,000  yards  of  2-ply  48s 

cotton?  Ans.  17.857  lb. 

(9)  What  is  the  length  of  18  pounds  of  8s  cotton? 

Ans.  120,960  yd. 

(10)  What  counts  must  be  twisted  with  a  14s  to  make  a 

2-ply  yarn  equal  in  weight  to  a  single  10s?  Ans.  35s 

(11)  What  weight  of  60s  will  be  required  to  produce 
120  pounds  of  2-ply  yarn  when  twisted  with  a  38s? 

Ans.  46.530  lb. 

(12)  What  counts  must  be  twisted  with  a  40s  to  make  a 

2-ply  yarn  equal  in  weight  to  a  24s  single?  Ans.  60s 

§9 


2 


YARN  CALCULATIONS,  COTTON 


9 


(13)  If  1  end  of  30s  and  1  end  of  10s  are  twisted,  what 
will  be  the  weight  of  each  in  50  pounds  of  the  ply  yarn? 

a  \  12.5  lb.  of  30s 
Ans-[37.5  lb.  of  10s 

(14)  A  given  length  of  2-ply  yarn  weighs  70  pounds  and 
is  composed  of  1  end  of  24s  and  1  end  of  9s;  what  weight 

of  each  is  there  in  the  ply  yarn?  A  \  19.089  lb.  of  24s 

Ans- 1 50.905  lb.  of  9s 


(15)  How  many  ends  are  there  in  a  32s  cotton  warp  200 

yards  long  and  weighing  15  pounds?  Ans.  2,016  ends 

(16)  If  76  pounds  of  30s  cotton  yarn  is  used  in  a  warp 
of  3,600  ends,  what  is  the  length  of  the  warp?  Ans.  532  yd. 

(17)  What  is  the  weight  of  yarn  in  a  cotton  warp  con¬ 

taining  2,800  ends,'  the  warp  being  60  yards  long  and  the 
counts  of  the  yarn  24s?  Ans.  83  lb. 

(18)  A  warp  is  arranged  8  ends  of  black  and  4  of  white; 
the  warp  is  160  yards  long  and  contains  1,800  ends  of  30s 
cotton.  What  is  the  weight  of  each  color  in  the  warp? 

a  f  3.809  lb.  white 

Ans>  17.618  lb.  black 

(19)  («)  A  cotton  warp  contains  4  pounds  of  18s  and 

has  1,200  ends;  how  long  is  it?  (b)  15  pounds  of  2-ply 
60s  cotton  is  made  into  a  warp  80  yards  long;  how  many 
ends  does  it  contain?  A  \  (a)  50.4  yd. 

Ans'l  {b)  4,725  ends 

(20)  Give  a  rule  for  finding  the  counts  of  a  ply  yarn 
composed  of  more  than  two  threads  of  single  yarn. 

(21)  State  how  to  find  the  weight  of  each  thread  in  a  ply 
yarn  when  the  total  weight  and  the  counts  of  the  separate 
threads  are  given. 

(22)  Two  threads  of  60s  and  one  of  40s  are  twisted;  what 

are  the  counts  of  the  ply  yarn?  Ans.  17.142s 

(23)  52^  pounds  of  20s  cotton  is  used  to  make  a  warp 
700  yards  long;  how  many  ends  are  there  in  the  warp? 

Ans.  1,260  ends 


§9 


YARN  CALCULATIONS,  COTTON 


3 


(24)  What  are  the  average  counts  of  the  yarn  in  a  warp, 
the  ends  arranged  34  ends  of  48s  and  1  end  of  2-ply  20s? 

Ans.  43.316s 

(25)  What  twist  would  be  inserted  in  an  ordinary  56s 

cotton  filling  yarn?  Ans.  24.319  turns  per  inch 


EXAMINATION  QUESTIONS 

(1)  What  counts  of  yarn  should  be  twisted  with  a  24s  to 
make  a  ply  yarn  equal  in  weight  to  an  8s  single?  Ans.  12s 

(2)  An  80s,  40s,  and  30s  are  twisted  together;  what  are 

the  counts  of  the  ply  yarn?  Ans.  14.117s 

(3)  How  many  yards  are  there  in  2  pounds  of  50s 

worsted?  Ans.  56,000  yd. 

(4)  Explain  what  is  meant  by  the  word  counts. 

(5)  State  what  is  meant  by  ply  yarns. 

(6)  State  how  many  yards  there  are  in  1  pound  of  yarn, 
the  counts  being  5s:  (a)  in  the  woolen  (run  and  cut)  systems; 
(b)  in  the  worsted  system. 

A  f  (a)  8,000  yd.  (woolen  run),  1,500  yd.  (woolen  cut) 
ns'i(£)  2,800  yd.  (worsted) 

(7)  If  368,000  yards  of  single  worsted  yarn  weighs  16 

pounds,  what  are  the  counts?  Ans.  41.071s 

(8)  What  counts  of  yarn  are  twisted  to  make  a  2-ply  38s 
worsted  yarn? 

(9)  What  is  the  weight  of  360,000  yards  of:  (a)  2-ply 
36s  worsted?  (b)  5-run  woolen? 

A  f  (a)  35.714  lb.  worsted 
ns,\(£)  45  lb.  woolen  run 


§10 


2 


YARN  CALCULATIONS, 


§10 


(10)  How  many  ounces  of  yarn  are  there  in  14,000  yards 

of  2-ply  36s  worsted?  Ans.  22.208  oz. 

(11)  What  is  the  length  of  18  pounds  of  8s  yarn:  (a)  in 

the  worsted  system?  ( b )  in  the  woolen  systems  (run  and 
cut)?  A  \  (a)  80,640  yd. 

ns‘t(£)  230,400  yd.  (run),  43,200  yd.  (cut) 

(12)  What  counts  must  be  twisted  with  a  single  14s  to 
produce  a  2-ply  thread  equal  in  weight  to  a  single  10s? 

Ans.  35s 

(13)  What  weight  of  60s  worsted  will  be  required  to 

produce  120  pounds  of  2-ply  yarn  when  twisted  with  a  38s 
worsted?  Ans.  46.530  lb. 

(14)  A  jack-spool  containing  40  ends  is  spooled  with  300 

yards  of  woolen  yarn  that  weighs  3  pounds;  what  run  is  the 
yarn?  Ans.  2.5-run 

(15)  A  worsted  warp  containing  2,800  ends  is  800  yards 
long  and  weighs  200  pounds;  what  is  the  size  of  the  yarn? 

Ans.  20s 

(16)  A  woolen  warp  is  400  yards  long  and  weighs 

80  pounds.  If  the  warp  is  composed  of  42"-run  yarn,  how 
many  ends  does  it  contain?  Ans.  1,440- ends 

(17)  A  worsted  warp  740  yards  long  contains  1,600  ends 
of  2-ply  36s;  what  is  the  weight  of  the  yarn  in  the  warp? 

Ans.  117.460  lb. 

(18)  A  warp  containing  2,772  ends  is  dressed  6  black, 
2  slate,  3  black,  1  slate,  12  black,  4  slate;  how  many  ends 
of  each  color  are  there  in  the  warp? 

Ans.  2,079  ends  of  black,  693  ends  of  slate 

(19)  A  woolen  warp  contains  864  ends  of  3-run  yarn  and 
weighs  85i  pounds;  what  is  the  length  of  the  warp? 

Ans.  475  yd. 

(20)  A  warp  contains  1,400  ends  of  2-ply  40s  worsted 

and  700  ends  of  10s;  what  is  the  average  number  of  the 
warp?  Ans.  15s 


§10 


WOOLEN  AND  WORSTED 


3 


(21)  If  2  yards  of  worsted  yarn  weighs  1  grain,  what 

is  the  size  of  the  yarn?  Ans.  25s 

(22)  If  1  yard  of  woolen  yarn  weighs  1.2  grains,  what 

run  is  the  yarn?  Ans.  3.645-run 

(23)  300,000  yards  of  woolen  yarn  weighs  50  pounds; 
what  are  the  counts  in  both  the  run  and  cut  systems? 

Ans.  3.75-run,  20-cut 

(24)  Give  a  rule  for  finding  the  counts  of  a  ply  yarn 
composed  of  more  than  two  threads  of  single  yarns. 

(25)  State  how  to  find  the  weight  of  each  thread  in  a  ply 
yarn  when  the  total  weight  and  the  counts  of  the  separate 
threads  are  given. 


YARN  CALCULATIONS, 
GENERAL 


EXAMINATION  QUESTIONS 

(1)  What  counts  of  yarn  should  be  twisted  with  a  24s 
cotton  to  make  a  ply  yarn  equal  in  weight  to  an  8s  single? 

Ans.  12s 


(2)  An  80s,  40s,  and  30s  worsted  are  twisted  together; 
what  are  the  counts  of  the  ply  yarn?  Ans.  14.117s 


(3)  How  many  yards  are  there  in  2  pounds  of  50s  cotton 

yarn?  Ans.  84,000  yd. 

* 

(4)  How  many  yards  are  there  in  13  pounds  of  3l-run 

woolen  yarn?  Ans.  70,200  yd. 

(5)  Explain  what  is  meant  by  the  word  counts. 


(6)  State  how  many  yards  there  are  in  1  pound  of  yarn 
in  each  of  the  following  systems,  the  counts  being  5s  single: 
woolen  (run  and  cut),  cotton,  spun  silk,  and  worsted. 

8,000  yd.  (woolen  run),  1,500  yd.  (woolen  cut) 
Ans. ( 4,200  yd.  (cotton),  4,200  yd.  (spun  silk) 

.2,800  yd.  (worsted) 


(7)  If  368,000  yards  of  single  worsted  yarn  weighs 

16  pounds,  what  are  the  counts?  Ans.  41.071s 

(8)  What  counts  of  single  yarns  are  twisted  to  make  a 
2-ply  38s  worsted  yarn? 


2 


YARN  CALCULATIONS,  GENERAL 


11 


(9)  Find  the  counts  in  the  woolen  (run  and  cut), 
worsted,  and  cotton  systems  equal  to  36s  spun  silk. 

«  j  18.9s  (woolen  run),  100.8s  (woolen  cut) 
ns'l54s  (worsted),  36s  (cotton) 

(10)  What  is  the  weight  of  360,000  yards  of:  ( a )  2-ply 
36s  worsted?  ( b )  2-ply  48s  cotton?  ( c )  5-run  woolen? 

(a)  35.714  lb. 

Ans.]  (b)  17.857  lb. 

Ac)  45  lb. 

(11)  How  many  ounces  of  yarn  are  there  in  14,000  yards 

of  2-ply  36s  worsted?  Ans.  22.208  oz. 

(12)  Change  2-ply  40s  worsted  to  2-ply  cotton  counts. 

Ans.  26fs 

(13)  What  is  the  length  of  18  pounds  of  8s  yarn:  (a)  in 
the  cotton  system?  ( b )  in  the  worsted  system?  (c)  in  the 


spun-silk  system? 


Ans. 


(a)  120,960  yd. 

(b)  80,640  yd. 
Ac)  120,960  yd. 


(14)  What  counts  must  be  twisted  with  single  14s 
worsted  to  produce  a  2-ply  yarn  equal  to  a  single  10s? 

Ans.  35s 


(15)  What  weight  of  60s  worsted  will  be  required  to  pro¬ 
duce  120  pounds  of  2-ply  yarn  when  twisted  with  a  38s 
worsted?  Ans.  46.530  lb. 


(16)  What  is  the  price  per  pound  of  a  ply  yarn  composed 
of  one  thread  of  36s  cotton  at  I4  cents  per  ounce  and  one 
thread  of  54s  worsted  at  6  cents  per  ounce?  Ans.  58  ct. 

(17)  What  counts  of  cotton  yarn  must  be  twisted  with 
a  40s  cotton  to  make  a  2-ply  yarn  equal  to  a  24s  single? 

Ans.  60s 

(18)  If  one  end  of  30s  and  one  of  10s  are  twisted,  what 
will  be  the  weight  of  each  yarn  in  50  pounds  of  the  ply  yarn? 

A  {12.5  lb.  of  30s 
Ans- 137.5  lb.  of  10s 


11 


YARN  CALCULATIONS,  GENERAL 


3 


(19)  •  A  given  length  of  2-ply  cotton  yarn  weighs 
70  pounds  and  is  composed  of  one  end  of  24s  and  one  end 
of  9s;  what  weight  of  each  is  there  in  the  ply  yarn? 

,  f  19.089  lb.  of  24s 
Ans- (50.905  lb.  of  9s 

(20)  300,000  yards  of  woolen  yarn  weighs  50  pounds; 
what  are  the  numbers  in  both  the  run  and  cut  systems? 

Ans.  3.75s  (run),  20s  (cut) 

(21)  A  60s  cotton,  40s  worsted,  and  60s  silk  are  twisted; 
what  are  the  counts  of  the  ply  thread  in  the  worsted  system? 

Ans.  21.176s 

(22)  How  many  ends  are  there  in  a  32s  cotton  warp 
200  yards  long  and  weighing  15  pounds?  Ans.  2,016  ends 

(23)  A  worsted  warp  containing  2,800  ends  is  800  yards 
long  and  weighs  200  pounds;  what  is  the  size  of  the  yarn? 

Ans.  20s 

(24)  If  76  pounds  of  30s  cotton  warp  yarn  is  used  in  a 
warp  of  3,600  ends,  what  is  the  length  of  the  warp? 

Ans.  532  yd. 

(25)  A  woolen  warp  is  400  yaids  long  and  weighs 

80  pounds;  if  the  warp  is  composed  of  42-run  yarn,  how 
many  ends  does  it  contain?  Ans.  1,440  ends 


CLOTH  CALCULATIONS, 

COTTON 


EXAMINATION  QUESTIONS 

(1)  How  many  ends  are  there  in  each  of  the  following- 
warps,  allowing  48  ends  for  selvages:  ( a )  Width  at  reed  80 
inches,  reed  20  dents  per  inch,  2  ends  per  dent?  (b)  Width 
at  reed  32  inches,  reed  40  dents  per  inch,  2  ends  per  dent? 

a  f  (a)  3,248  ends 
s‘l(£)  2,608  ends 

(2)  A  warp  contains  3,800  ends  and  is  woven  in  a  reed 
containing  30  dents  per  inch,  the  ends  being  drawn  in  4  per 
dent;  what  is  the  width  at  the  reed,  neglecting  selvages? 

Ans.  31f  in. 

(3)  What  is  the  weight  of  yarn  in  a  warp  28  inches  wide 

at  the  reed  including  selvages,  drawn  6  per  dent  in  a  reed 
having  12  dents  per  inch,  the  length  of  the  warp  being 
50  yards  and  the  counts  of  the  yarn  60s?  Ans.  2  lb. 

(4)  What  is  the  weight  of  yarn  in  a  warp  containing 

2,800  ends  including  selvages,  the  warp  being  60  yards  long 
and  the  counts  of  the  yarn  24s?  Ans.  8i  lb. 

(5)  A  cut  of  cloth  made  with  36s  filling  is  50  yards  long, 
32  inches  wide  at  the  reed,  including  selvages,  and  contains 
56  picks  per  inch;  what  is  the  weight  of  the  filling? 

Ans.  2.962  lb. 

(6)  (a)  A  warp  is  arranged  8  ends  of  black  and  4  of 
white.  It  is  160  yards  long  and  contains  1,800  ends  of  30s; 

§  12 


2 


CLOTH  CALCULATIONS,  COTTON 


12 


what  is  the  weight  of  each  color  in  the  warp?  (b)  What 
weight  of  filling  will  be  required  if  the  above  piece  is  woven 
30  inches  wide  at  the  reed,  including  selvages,  with  64  picks 
per  inch,  the  warp  yielding  150  yards  of  cloth?  The  counts 
of  the  filling  are  the  same  as  the  warp.  ( c )  What  is  the 
weight  per  yard  in  ounces  of  the  above  cloth,  allowing  10  per 
cent,  for  size  on  warp  yarns? 

(a)  3.809  lb.  of  white,  7.618  lb.  of  black 
Ans.{  {b)  11.428  lb. 

(c)  2.5  oz. 


(7)  (a)  A  warp  contains  4  pounds  of  18s  and  has 

1,200  ends;  how  long  is  it?  (b)  15  pounds  of  2-ply  60s  is 
made  into  a  warp  80  yards  long;  how  many  ends  does  it 
contain?  .  f  (a)  50.4  yd. 

Ans’t(£)  4,725  ends 

(8)  What  is  the  cost  of  the  yarn  in  a  piece  of  cloth  made 
as  follows:  70  ends  per  inch,  in  the  reed,  of  35s  at  20  cents 
per  pound,  72  picks  per  inch  of  40s  at  22  cents  per  pound? 
The  piece  is  woven  55  yards  long  from  58  yards  of  warp  and 
is  30  inches  wide  at  the  reed,  including  selvages. 

Ans.  $1,606 

(9)  52i  pounds  of  20s  is  used  to  make  a  warp  700  yards 

long:  (a)  How  many  ends  are  there  in  the  warp?  (b)  If 
the  yarn  is  21  inches  wide  at  the  reed,  including  selvages, 
and  the  reed  contains  20  dents  per  inch,  the  selvages  being 
reeded  like  the  body  of  the  warp,  how  many  ends  are  there 
to  the  dent?  A  \  (a)  1,260  ends 

11S’1(£)  3  ends  per  dent 

(10)  A  warp  pattern  is  as  follows:  4  ends  of  blue,  16  ends 
of  white,  4  ends  of  red.  The  width  at  the  reed  is  32  inches, 
not  including  selvages,  and  the  warp  is  drawn  2  ends  per 
dent  in  a  reed  containing  30  dents  per  inch.  How  many 
ends  of  each  color  will  be  required  for  the  body  of  the 
warp?  Ans.  320  ends  red,  320  ends  blue,  1,280  ends  white 

(11)  What  will  be  the  weight  of  filling  in:  («)  56  yards 
of  cloth  29  inches  wide  at  the  reed,  including  selvages,  78 
picks  per  inch,  using  60s  filling?  {b)  100  yards  of  cloth 


§12 


CLOTH  CALCULATIONS,  COTTON 


3 


32  inches  wide  at  the  reed  including  selvages,  36  picks  per 
inch,  using  16s  filling?  A  f  (a)  2.513  lb. 

Ans‘l(£)  8.5711b. 

(12)  A  cloth  is  woven  100  yards  long  from  104  yards  of 
warp  29  inches  wide  at  the  reed,  including  selvages,  64  picks 
per  inch  of  36s  filling,  and  64  ends  per  inch  of  27s  warp; 
what  will  be  the  weight  per  yard  in  ounces  of  above  cloth, 
allowing  10  per  cent,  for  size  on  warp  yarn?' 

Ans.  2.47  oz. 


(13)  What  length  of  cloth  will  100  pounds  of  30s  supply 
with  filling  if  the  cloth  is  to  be  29  inches  wide  at  the  reed 
including  selvages,  and  to  contain  72  picks  per  inch? 

Ans.  1,206.896  yd. 

(14)  What  is  the  weight  per  yard,  in  ounces,  of  a  cloth 

56  inches  wide  including  selvages,  if  1  square  inch  weighs 
2.625  grains?  Ans.  12.096  oz. 

(15)  A  cloth  28  inches  wide  at  the  reed  including  sel¬ 

vages  is  made  with  56  ends  per  inch  in  the  reed,  and  56 
picks  per  inch  of  30s.  What  is  the  weight  per  yard  in 
ounces,  adding  7  per  cent,  to  warp  yarn  for  size  and  con¬ 
traction  in  weaving?  Ans.  2.048  oz. 


(16)  A  warp  pattern  is  drawn  in  as  follows: 


Dents 

16 

4 

2 

4 

26 


Ends 

32 

16 

4 

16 

68 


(a)  How  many  patterns  will  there  be  in  a  warp  30  inches 
wide  at  the  reed  excluding  selvages,  using  a  34s  reed? 
( b )  What  will  be  the  total  number  of  ends  required  for  the 
body  of  the  warp?  A  \(a)  39  patterns,  6  dents  over 

Ans,l(£)  2,664  ends 


(17)  A  piece  of  cloth  4  inches  square  weighs  14  grains; 
if  the  cloth  is  32  inches  wide  including  selvages,  what  are 
the  yards  per  pound?  Ans.  6.944  yd. 


4 


CLOTH  CALCULATIONS,  COTTON 


§12 


(18)  What  would  be  the  width  in  the  reed  of  a  64-sley 
cloth  containing  1,800  ends  in  the  body  of  the  cloth  drawn 
2  per  dent,  and  48  ends  drawn  4  per  dent  for  selvages? 

Ans.  30.476  in. 

(19)  What  reed  will  be  required  to  weave  a  50-sley  cloth, 

if  drawn  2  ends  per  dent?  Ans.  23.275s 

(20)  A  cloth  is  of  the  following  construction:  70  X  76 

—  38  in.  wide  including  selvages  —  70s  warp  —  12  yd.;  what 
counts  of  filling  must  be  used?  Ans.  118s  (about) 


CLOTH  CALCULATIONS, 
WOOLEN  AND  WORSTED 


EXAMINATION  QUESTIONS 

(1)  A  worsted  cloth  is  28  inches  wide  and  1  square  inch 
weighs  2.6  grains;  what  is  the  weight  of  the  cloth  per  yard? 

Ans.  5.99  oz. 


(2)  A  warp  containing  2,600  ends  is  to  be  drawn  through 
the  harnesses  according  to  the  draft  shown  in  Fig.  1;  how 
many  heddles  are  required  for  each  harness? 


Ans. 


780  on  first  harness 
1,040  on  second  harness 
520  on  third  harness 
260  on  fourth  harness 


(3)  A  warp  containing  2,772  ends  is  dressed  6  black, 
2  slate,  3  black,  1  slate,  12  black,  4  slate;  how  many  ends  of 
each  color  are  there  in  the  warp? 

*  ,  [2,079  ends  of  black 

ns*  1 693  ends  of  slate 


(4)  The  above  warp  is  composed  of  2 /40s  worsted  yarn 
and  is  560  yards  long;  what  is  the  weight  of  each  color  of 
yarn? 

A  [103.95  lb.  black 
134.65  lb.  slate 

(5)  A  warp  contains  1,400  ends  of  2/40s  worsted  and 
700  ends  of  10s;  what  is  the  average  number  of  the  warp? 

Ans.  15s 


2 


CLOTH  CALCULATIONS, 


13 


(6)  A  warp  contains  2,520  ends  including  selvage  ends 
and  is  reeded  5  per  dent;  if  there  are  70  ends  per  inch  in  the 
reed:  (a)  what  reed  should  be  used?  (b)  what  is  the  width 
of  the  goods  in  the  loom? 

a  f  (a)  14s  reed 
Ans-1(*)  36  in. 

(7)  If  50  inches  of  worsted  yarn  weighs  1  grain,  what  is 

the  size  of  the  yarn?  Ans.  17.361s 

(8)  A  piece  of  cloth  contains  1,960  ends  of  20s  worsted 

in  the  warp;  the  cloth  is  30  inches  wide  and  contains  44  picks 
per  inch  of  4-run  woolen  yarn;  what  is  the  weight  of  the 
cloth  per  yard?  Ans.  6.1  oz. 

(9)  If  30  inches  of  woolen  yarn  weighs  1.2  grains,  what 

run  is  the  yarn?  Ans.  3.038-run 

(10)  A  cloth  is  woven  62  inches  wide  in  the  reed;  the 

woven  cloth  has  28  picks  per  inch  of  3i-run  yarn;  in  a  cer¬ 
tain  length  of  cloth  there  are  14  pounds  of  filling;  what  is  the 
length?  Ans.  45.16  yd. 

(11)  One  square  inch  of  cloth  weighs  2.8  grains;  what  is 
the  weight  of  a  yard  if  the  cloth  is  56  inches  wide? 

Ans.  12.902  oz. 

(12)  A  certain  piece  of  cloth  has  shrunk  20  per  cent,  from 

reed  to  finished  goods  and  lost  10  per  cent,  in  finishing;  the 
filling  in  the  finished  cloth  is  14.4s  worsted;  what  was  the 
original  size?  Ans.  16s 

(13)  A  piece  of  cloth  is  36.96  inches  wide  finished  and 

has  shrunk  16  per  cent,  from  reed  to  finished  cloth;  what  was 
the  width  in  the  loom?  Ans.  44  in- 

(14)  A  warp  contains  1,120  ends  of  22-run  yarn  and  takes 

up  4  per  cent,  in  weaving;  what  is  the  weight  of  warp  in 
1  yard  of  cloth  from  the  loom?  Ans.  4.65  oz. 

(15)  If  a  piece  of  cloth  is  34  inches  wide  and  contains 

28  picks  per  inch  of  10s  worsted,  what  is  the  weight  of  filling 
in  48  yards  of  cloth?  Ans.  8.160  lb. 


§13  WOOLEN  AND  WORSTED  3 

(16)  What  is  the  weight  of  a  yard  of  cloth  composed  of 

960  ends  and  32  picks  per  inch  of  3.2-run  yarn  if  it  is  30  inches 
wide?  Ans.  6  oz. 

(17)  If  the  above  cloth  has  shrunk  20  per  cent.,  both  warp 
and  filling  ways,  what  was  the  original  size  of  the  yarn  and 
what  was  the  width  in  the  reed,  supposing  the  finished  width 
given  in  the  previous  example  to  be  outside  of  selvages? 
Neglect  loss  in  finishing. 

A  J  4-run 
Ans*  1 37.5  in. 

(18)  A  warp  contains  2,592  ends  and  is  reeded  2  per  dent, 
the  reed  being  an  18s;  what  is  the  width  in  the  loom? 

Ans.  72  in. 

(19)  What  is  the  weight  of  the  filling  in  a  yard  of  goods 

44  inches  wide  and  containing  44  picks  per  inch  of  14.65s 
worsted  filling?  Ans.  3.775  oz. 

(20)  A  warp  is  reeded  3  per  dent  and  is  37  inches  wide 

in  the  loom;  what  is  the  number  of  ends  in  the  warp  if  an 
11s  reed  is  used?  Ans.  1,221  ends 


EXAMINATION  QUESTIONS 

(1)  What  are  the  objects  of  drafting? 

(2)  Why  is  more  draft  produced  by  a  set  of  metallic 
rolls  than  by  a  set  of  common  rolls  that  correspond  in 
diameter  with  the  metallic  rolls  and  that  are  geared  in  the 
same  manner? 

(o)  If  the  number  of  ends  put  up  at  the  back  of  a  certain 
machine  and  drawn  into  one,  the  weight  per  yard  of  each 
end  at  the  back,  and  the  weight  per  yard  of  the  end 
delivered  at  the  front  are  known,  how  could  the  draft  of  the 
machine  be  obtained? 

(4)  Explain  the  terms:  (a)  draft  change  gear;  (b)  crown 
gear. 

(5)  Give  the  rule  for  finding  the  required  draft  gear  to 
give  a  certain  draft  when  the  constant  is  known,  the  constant 
being  a  constant  dividend. 

(6)  What  is  the  effect  of  placing  a  larger  draft  gear  on 
machines  geared  similarly  to  Figs.  4,  5,  and  6? 

(7)  Explain  the  terms:  (a)  constant  factor;  (b)  constant 
dividend. 

(8)  What  is  meant  by  saying  that  a  machine  has  a 
draft  of  6? 

(9)  Explain  the  term  break  draft. 

(10)  Give  the  rule  for  finding  the  draft  constant  of  a  set 
of  rolls  connected  by  gears. 


2 


DRAFT  CALCULATIONS 


§15 


(11)  Find  the  break  draft  for  the  rolls  and  gears  shown 

in  Fig.  5.  Ans.  2.915 

(12)  Find  the  total  draft  of  a  machine  that  has  a  draft  of 

1.4  between  third  and  back  rolls;  a  draft  of  1.5  between 
second  and  third  rolls,  and  a  draft  of  3.2  between  front  and 
second  rolls.  Ans.  6.72 

(13)  If  the  feed-roll  and  delivery  roll  of  a  machine  make 

the  same  number  of  revolutions  per  minute,  what  will  be  the 
draft  if  the  feed-roll  is  If  inches  in  diameter  and  the  delivery 
roll  If  inches  in  diameter?  Ans.  1.555 

(14)  Find  the  total  draft  for  Fig.  5,  supposing  that 

metallic  rolls  are  used  and  a  60-grain  sliver  is  passing  into 
the  machine.  Ans.  7.04 

(15)  Find  the  total  draft  for  the  set  of  rolls  shown  in 

Fig.  6.  Ans.  6.006 

(16)  Find  the  total  draft  for  the  set  of  rolls  shown  in 

Fig.  5.  Ans.  6.459 

(17)  Find  the  constant  for  Fig.  5,  the  gear,g  being  the 

change  gear.  Ans.  523.25 

(18)  Find  the  constant  for  Fig.  6,  the  gear^'  being  the 

change  gear.  Ans.  468.518 

(19)  Using  the  constant  found  in  answer  to  question  18, 
what  change  gear  will  be  required  to  give  a  draft  of  7? 

Ans.  67,  gear 

(20)  If  a  draft  gear  of  68  teeth  gives  a  draft  of  6.04  on  a 
certain  machine,  what  draft  gear  will  be  required  to  give  a 
draft  of  6.4,  the  draft  gear  being  a  driver?  Ans.  64,  gear 


READING  TEXTILE 
DRAWINGS 


EXAMINATION  QUESTIONS 

(1)  Name  two  uses  of  dotted  lines. 

(2)  What  is  an  assembly  drawing? 

(3)  (a)  What  is  a  perspective  drawing?  ( b )  What  is  the 
principal  difference  between  a  perspective  drawing  and  a 
photographic  reproduction  of  the  same  object  from  the  same 
point  of  view? 

(4)  What  is  meant  when  it  is  said  that  a  certain  drawing 
is  a  right-side  elevation  of  an  object? 

(5)  For  what  are  broken  lines  chiefly  used? 

(6)  In  a  drawing  of  a  section,  what  is  done  by  the  drafts¬ 
man  to  enable  it  to  be  recognized  as  a  sectional  view  and  not 
one  of  the  outside  surfaces  of  the  object? 

(7)  Show,  by  sketches,  how  cast  iron,  wrought  iron,  steel, 
and  brass  are  indicated  by  section  lines. 

(8)  When  it  is  desired  to  draw  a  long  piece,  such  as  a 
shaft,  in  a  limited  space,  how  can  the  drawing  be  made  with¬ 
out  reducing  the  size  of  the  object  to  such  an  extent  as  to 
make  the  diameter  appear  so  small  that  details  cannot  be 
shown  plainly? 

(9)  Give  a  sketch  showing  the  conventional  method  of 
representing  a  screw  thread  that  is  largely  used. 

I  91 


2  READING  TEXTILE  DRAWINGS  §91 

(10)  How  are  sections  of  an  object  shown  that  is  too 
small  or  thin  to  be  readily  sectioned? 

(11)  Why  is  Fig.  15  (c)  preferable  to  Fig.  15  (£)? 

(12)  What  is  a  diagrammatic  drawing? 

(13)  State  briefly  the  use  of  shade  lines. 

(14)  In  the  object  shown  in  Fig.  I,  (a)  how  many  separate 

pieces  are  shown?  (5)  what 
different  materials  are  indi¬ 
cated? 

(15)  Make  a  sketch  show¬ 
ing  a  conventional  method 
of  representing  a  spring. 

(16)  How  can  it  be  deter¬ 
mined  if  a  screw  has  a  right- 
hand  thread? 

(17)  Why  are  reference 
letters  placed  on  drawings? 

(18)  What  can  be  said  in 
regard  to  reference  letters  referring  to  a  piece  that  is  shown 
in  several  drawings? 

(19)  What  is  meant  by  a  partial  section? 

(20)  If  a  drawing  of  an  object  is  made  to  an  eighth  scale 
and  the  object  is  36  inches  long,  what  will  be  the  length  of 
the  drawing? 


INDEX 


Note.— All  items  in  this  index  refer  first  to  the  section  and  then  to  the  page  of  the  sec¬ 
tion.  Thus,  “Addition  1  6”  means  that  addition  will  be  found  on  page  6  of  section  1. 


A 

Sec.  Page 

Sec.  Page 

Abstract  number . 

1 

1 

Average  number  of  the  warp  .  .  . 

10 

23 

Acute  angle . 

6 

2 

yarns . 

9 

15 

Addition . 

1 

6 

“  “  **  “ 

11 

33 

of  compound  quantities  . 

5 

12 

sley . 

12 

23 

“  decimals . 

2 

21 

Avoirdupois  weight . 

5 

3 

“  fractions . 

2 

9 

Axis  of  a  sphere . 

6 

15 

“  Proof  of . 

1 

11 

Rule  for . 

1 

10 

B 

Sign  of . 

1 

6 

Back  rolls . 

15 

6 

Aggregation,  Signs  of . 

3 

19 

Backlash . 

7 

29 

Altitude  of  a  parallelogram  .... 

6 

5 

Base . 

3 

2 

“  pyramid  or  cone  .  .  . 

6 

14 

“  of  a  prism . 

6 

11 

“  “trapezoid . 

6 

5 

rate,  and  percentage . 

3 

4 

“  “  triangle . 

6 

2 

Beam  calculations  , . 

9 

13 

American  warp  yarns.  Breaking 

“  “ 

10 

18 

weight  of . 

9 

20 

Beamed  yarns . 

11 

30 

Amount . 

S 

3 

Beams,  Loom . 

9 

13 

tl 

3 

7 

“  Section . 

9 

13 

Angles . 

6 

2 

Belt,  Crossed . 

7 

20 

Angular  measure . 

5 

7 

"  fastenings . 

7 

11 

Animal  fibers . 

11 

1 

“  Open . 

7 

20 

Annular  gear . 

7 

34 

“  Quarter-turn . 

7 

22 

Antecedent . 

3 

11 

“  shipper . 

7 

18 

Apothecaries’ fluid  measure  .  .  .  . 

5 

4 

Belting . 

7 

10 

weight . 

5 

4 

Belts,  Care  of . 

7 

12 

Arabic  notation . 

1 

2 

“  Cotton . 

7 

10 

Arc  of  a  circle . 

6 

8 

“  Horsepower  transmitted  by 

8 

7 

Area . 

6 

2 

“  Leather . 

7 

10 

“  Measures  of . 

5 

6 

"  Length  of . 

8 

6 

“  of  a  circle . 

6 

9 

Rubber . 

7 

10 

“  “  “  cylinder . 

6 

12 

“  Rules  applying  to . 

8 

6 

“  M  prism . 

6 

11 

“  Single  and  double . 

7 

11 

“  “  “  regular  polygon  .... 

6 

7 

Bevel  gears . 

7 

30 

.  sphere  . 

6 

15 

Binder  pulley . 

7 

17 

“  “trapezium. . 

6 

6 

Board  measure . 

6 

16 

“  “  “  trapezoid . 

6 

5 

Body,  Solid . 

6 

10 

“  “  triangle . 

6 

3 

Borrowing  numbers . 

1 

13 

Arithmetic . 

1 

1 

Box  couplings . 

7 

2 

Ascending,  Reduction . 

5 

1 

Braces  . 

3 

19 

Assembly  and  detail  drawings  .  . 

91 

17 

Brackets . 

3 

19 

Average  counts . 

11 

33 

“  Wall . 

7 

7 

“  “ 

12 

21 

Break  draft . 

15 

17 

XI 


Xll 


INDEX 


Sec.  Page 

Sec.  Page 

Breaking-  weight  of  warp  yarns  .  . 

9 

19 

Circle,  Radius  of  a . 

6 

8 

Breaks  in  drawings . 

91 

25 

“  Rule  to  find  area  of . 

6 

9 

Broken-and-dotted  lines . 

91 

22 

“  “  “  circumference 

“  line . 

91 

22 

of . 

6 

8 

Bushings,  Pulley . 

7 

15 

Circular  measure . 

5 

7 

“  pitch  of  a  gear . 

7 

29 

C 

Circumference  of  a  circle . 

6 

8 

Calculating  draft  for  common  rolls 

15 

12 

Clamp  coupling . 

7 

3 

production  and  draft  . 

15 

22 

Classes  of  shafting . 

7 

1 

Calculation  of  cost  of  ply  yarns  .  . 

11 

25 

Cloth  calculations,  Cotton  .... 

12 

1 

Calculations,  Beam . 

9 

13 

Importance  of 

12 

1 

It  i  i 

10 

18 

“  “  **  “ 

13 

1 

Cotton-cloth . 

12 

1 

Woolen  and 

“  yarn  . 

9  * 

1 

worsted  .  . 

13 

1 

“  Draft . 

15 

1 

“  Counts  of . 

12 

2 

for  filling  yarn  .  .  . 

12 

10 

“  Double-width . 

13 

3 

warp  yarns  .  .  . 

12 

5 

“  from  loom.  Weight  of .  .  .  . 

13 

24 

General  yarn  .... 

11 

1 

“  Ounce . 

12 

2 

Harness . 

12 

3 

“  Pick . 

12 

2 

(1  “ 

13 

4 

“  production,  C  al  cul  a  ti  on  s 

Importance  of  cloth  . 

12 

1 

necessary  for . 

12 

1 

“  “  “  “ 

13 

1 

“  production,  C  alcul  at  ion  s 

Mechanical . 

8 

1 

necessary  for . 

13 

1 

necessary  for  cloth 

“  samples,  F iguring  particu- 

production  .... 

12 

1 

lars  from  .... 

12 

16 

necessary  for  cloth 

F  i  g  u  r  ing  particu- 

production  .... 

13 

1 

lars  from  .... 

13 

21 

of  cotton  ply  yarns  . 

11 

20 

Obtaining  particu- 

“  ply  yarns . 

9 

8 

lars  of  . 

12 

24 

it  it  it  it 

10 

8 

“  Single-width . 

13 

3 

“single  cotton 

“  Weight  of . 

13 

3 

yarns . 

11 

3 

“  “  “ 

13 

21 

"  single  worsted 

“  Yard . 

12 

2 

yarns . 

10 

2 

Clutch  coupling . 

7 

4 

“  single  yarns  .  .  . 

9 

2 

Clutches . 

7 

2 

“  Reed . 

13 

5 

Friction . 

7 

5 

Woolen  and  worsted 

Cogs . 

7 

28 

cloth  . 

13 

1 

Cold-rolled  shafting,  Rule  to  find 

Woolen  and  worsted 

the  size  of . 

8 

1 

yarn . 

10 

1 

Colors,  Repeat  of . 

9 

17 

Cam-bowl . 

7 

37 

Common  denominator  of  fractions 

2 

8 

Cams . 

7 

35 

divisor.  Greatest  .... 

1 

29 

Cancelation . 

1 

33 

multiple . 

1 

31 

Canceling . 

1 

33 

rolls.  Calculating  draft 

Care  of  belts . 

7 

12 

for . 

15 

12 

Carrier  gear . 

7 

35 

Drafting  with  .  . 

15 

5 

Center  of  a  circle . 

6 

8 

Complex  fraction . 

2 

3 

Chain,  Engineer’s . 

5 

6 

Composite  number . 

i 

27 

“  Gunter’s . 

5 

6 

Compound  interest . 

3 

8 

Change  gear . 

8 

14 

levers . 

7 

38 

<1  <4 

15 

25 

numbers  . 

5 

1 

Characters . 

1 

2 

“  “  To  add  ... 

5 

13 

Chord  of  a  circle . 

6 

8 

quantities,  Addition  of 

5 

12 

Cipher . 

1 

2 

Division  of 

5 

16 

Circle . 

6 

8 

Multiplica- 

“  of  a  gear,  Pitch . 

7 

28 

tion  of  .  . 

5 

14 

INDEX 


xm 


Sec. 


Compound  quantities,  Reduction 

of  ...  .  5 

Subtraction 
of  ...  .  5 

Concentric  gears  .  7 

Concrete  number .  1 

Cone  and  pyramid .  6 

“  or  pyramid,  Altitude  of  a  .  .  6 

Frustum  of  a  .  .  6 

Cones,  Speed .  7 

Consequent  .  3 

Constant  dividend .  8 

dividends . 15 

factor  .  8 

factors . 15 

Rule  to  find  a .  8 

Constants .  8 

“  15 

Construction  of  fabrics . 13 

Contact,  Frictional .  7 

Contraction  during  weaving  ...  12 

in  weaving . 13 

Cost  of  ply  yarns,  Calculation  of  .  11 

Cotton  belts .  7 

“  cloth  calculations . 12 

ply  yarns.  Calculations  of  11 

“  roving  and  yarn,  Sizing  .  .  11 

warp  yarn.  Breaking 

weight  of .  9 

“  yarn  calculations .  9 

“  yarns  . 11 

Countershafts .  7 

“  7 

on  speed,  Effect  of  8 

Counts  and  weight.  Proof  of  .  .  .  13 

“  Average  . 11 

“  “  . .  12 

“  Equivalent . 11 

“  Folded  yarns  of  different  .  9 


“  .  .  10 

“  .  .  11 

the  same  9 

“  . .  “  10 

“  “  “  “  “  11 

Method  of  determining  the  11 


“  of  cloth . 12 

“  filling,  to  preserve  yards 

per  pound . 12 

“  the  warp  and  filling  .  .  13 

Couplet .  3 

Couplings .  7 

Cross-section . 91 

Crossed  belt .  7 

Crown-faced  pulley .  7 

Cube,  A .  6 

“  Definition  of  a .  5 


Sec.  Page 


Cube  of  a  number .  4  1 

“  root .  4  3 

“  “  4  10 

“  “  Table  method  of  ex¬ 
tracting  .  4  16 

Cubes  and  squares .  4  17 

Cubic  measure .  5  6 

Curved  line,  A .  6  1 

“  surface,  A .  6  2 

Cut  scale . 11  9 

“  system  of  numbering  woolen 

yarns . 10  14 

numbering  woolen 

yarns . 11  9 

Cylinder,  Surface  area  of  a  ...  .  6  12 

“  The .  6  12 

Volume  of  a .  6  13 

1) 

Dark  surface . 91  23 

Decagon .  6  7 

Decimal,  Reducing  a  fraction  to  a  2  29 

to  a  fraction,  Reducing  a  2  30 

Decimals .  2  18 

Addition  of .  2  21 

Division  of .  2  24 

Multiplication  of  ...  .  2  23 

Rule  for  the  division  of  .  2  27 

Subtraction  of .  2  22 

Definitions .  1  1 

Mechanical .  7  1 

of  mensuration  ....  6  1 

“  percentage .  3  1 

Delivery  rolls . 15  6 

Denominate  numbers .  5  1 

Denominator .  2  1 

of  fractions.  Common  2  8 

Dent . 13  6 

Descending,  Reduction .  5  1 

Detail  and  assembly  drawings  .  .  91  17 

Determining  the  counts,  Method  of  11  2 

“  diameter  of  yarns  11  29 

Diagrammatic  views . 91  29 

Diameter  of  a  circle .  6  8 

“  sphere  .......  6  15 

“  gear-blank .  8  12 

“  yarns . 11  29 

Diameters  of  pulleys,  Ratio  of 

speeds  to  ...  .  8  5 

“  rolls  on  draft,  Ef¬ 
fect  of  the  ....  15  12 

Diametral  pitch  of  a  gear .  7  29 

Difference .  1  12 

“  3  3 

Different  counts.  Folded  yarns  of  9  9 

Differential  screw .  7  35 


Page 

9 

13 

32 

1 

13 

14 

14 

18 

11 

15 

28 

14 

28 

15 

14 

28 

2 

9 

8 

9 

25 

10 

1 

20 

4 

19 

1 

2 

2 

21 

4 

22 

33 

21 

18 

9 

9 

22 

8 

8 

20 

2 

2 

22 

22 

11 

2 

15 

20 

14 

11 

6 


XIV 


INDEX 


Sec. 

Page 

Sec. 

Page 

Digits . 

i 

2 

Drawings,  Kinds  of . 

91 

2 

Dimension  lines . 

91 

27 

Lines  used  on . 

91 

22 

Direct  proportion . 

3 

13 

Mechanical . 

91 

3 

“  ratio . 

3 

11 

Reading  textile  .... 

91 

1 

Direction  of  pulley  rotation  .... 

7 

20 

Driven  and  driving  gears . 

15 

11 

Distance  between  hangers  .... 

7 

7 

Drives,  Rope  . 

7 

24* 

Dividend . 

1 

22 

Driving  and  driven  gears . 

15 

11 

Constant  . 

8 

15 

Drum,  A . 

7 

16 

Dividends,  Constant  . 

15 

28 

Dry  measure . 

5 

5 

Division . 

1 

22 

of  compound  quantities 

5 

16 

E 

“  decimals . 

2 

24 

Eccentric  gears . 

7 

32 

Rule  for  the 

2 

27 

Effect  of  countershafts  on  speed 

8 

4 

“  fractions . 

2 

15 

“  diameter  of  rolls  on 

Proof  of . 

1 

25 

draft . 

15 

12 

Rule  for . 

1 

25 

“  gears  and  rolls  on 

Sign  of . 

1 

22 

draft . 

15 

14 

Divisor  . 

1 

22 

size  of  gears  on  draft  .  . 

15 

13 

“  Greatest  common  .... 

1 

29 

Elements,  Machine . 

7 

1 

Dodecagon  . 

6 

7 

Elevations . 

91 

6 

Dot-and  dash  lines  . 

91 

22 

Elliptic  gears  . 

7 

34 

Dotted  line . 

91 

22 

End . 

13 

2 

Double  and  single  belts . 

7 

11 

Ends  and  reed  per  dent . 

13 

24 

"  width  cloth . 

13 

3 

“  Definition  of . 

12 

2 

yarns  . 

10 

7 

“  in  a  warp,  Finding  number 

Doubling . 

15 

3 

of . 

12 

5 

*  4 

15 

30 

“  the  warp,  Number  of  .  . 

13 

22 

Draft  and  production,  Calculating 

15 

22 

Engineer’s  chain . 

5 

6 

“  Break . 

15 

17 

Equality,  Sign  of . 

1 

6 

“  calculations . 

1 

Equilateral  triangle . 

6 

2 

“  Combined  effect  of  gears  and 

Equivalent  counts . 

11 

18 

rolls  on . 

15 

14 

Even  number . 

1 

28 

“  Drawing-in . 

13 

4 

Evolution . 

4 

3 

“  Effect  of  diameters  of  rolls 

Exponent  of  a  number . 

4 

1 

on . 

15 

12 

Extension  lines . 

91 

27 

“  Effect  of  size  of  gears  on  .  . 

15 

13 

Extracting  square  and  cube 

“  for  common  rolls,  Calcu- 

roots . 

4 

16 

lating . 

15 

12 

gears  . 

15 

25 

F 

“  Harness  . 

12 

3 

Fabrics,  Construction  of . 

13 

2 

13 

4 

Factor,  Constant . 

8 

14 

“  Methods  of  finding . 

15 

7 

Prime . 

1 

27 

Drafting . 

15 

1 

Factors . 

1 

27 

with  common  rolls  .  .  . 

15 

5 

“  Constant . 

15 

28 

“  special  reference  to 

“  Rational  and  irrational  .  . 

4 

8 

the  mill . 

15 

25 

Rule  to  find  prime  .... 

1 

28 

Drafts,  Irregular  reed . 

12 

12 

Fancy  patterns . 

13 

14 

“  Table  of . 

15 

24 

“  warps  . 

9 

16 

Drawing,  Definition  of  . 

91 

1 

“  “ 

10 

21 

Fundamental  principles 

4  (  4  4 

11 

35 

of . 

91 

3 

Fast  couplings . 

7 

2 

Scales  for . 

91 

26 

Fastenings,  Belt . 

7 

11 

in  draft . 

4 

Feed-rolls . 

15 

G 

Drawings,  Breaks  in . 

91 

25 

Fibers,  Animal  . 

11 

1 

Detail  and  assembly  .  . 

91 

17 

“  Vegetable . 

11 

1 

Freehand  . 

91 

2 

Figure,  A  plane . 

6 

2 

INDEX 


xv 


Sec.  Page 

G 

Sec. 

Page 

Figures  . 

1 

2 

Gauge  points . 

10 

6 

Significant  . 

4 

16 

Cl  Cl 

11 

13 

Figuring  particulars  from  cloth 

Gear,  Annular . 

7 

34 

samples . 

12 

16 

11  blank,  Diameter  of . 

8 

12: 

particulars  from  cloth 

“  blanks,  Sizing . 

8 

12: 

samples . 

13 

21 

“  Carrier  . 

7 

35 

Filling . 

11 

35 

“  Change  . 

8 

14 

( « 

13 

2 

c  c  c  c 

15 

25 

“  and  warp.  Counts  of  the  ,  . 

13 

22 

“  Idle . 

7 

35 

“  Finding  hanks  of . 

12 

12 

“  Intermediate . 

7 

35 

II  “  ti  << 

12 

20 

Pitch  circle  of  a . 

7 

28 

“  to  preserve  yards  per 

“  “  of  a  . 

7 

29 

pound,  Counts  of  .  .  .  . 

12 

22 

Ratchet . 

7 

31 

“  Weight  of . 

13 

23 

Gearing . 

7 

27 

“  yarn.  Calculations  for  .  .  . 

12 

10 

of  rolls . 

15 

9 

Finishing,  Shrinkage  in . 

13 

U 

Gears  and  rolls  on  draft,  Effect  of  . 

15 

14 

First  or  prime  mover . 

7 

2 

“  Bevel . 

7 

30 

Flange  coupling . 

7 

3 

“  Concentric . 

7 

32 

“  pulleys . 

7 

16 

“  Draft . 

15 

25 

Flat  pulley . 

7 

14 

“  Driving  and  driven . 

15 

11 

Floor  stand . 

7 

9 

“  Eccentric . 

7 

32 

Fluid  measure.  Apothecaries’  .  , 

5 

4 

Effect  of  size  of,  on  draft  .  . 

15 

13 

Folded  yarns . 

9 

7 

“  Elliptical . 

7 

34 

i  <  c  c 

10 

7 

Mangle . 

7 

32 

11 

20 

“  Miter . 

7 

31 

of  different  counts  .  . 

9 

9 

Mortise . 

7 

28 

<«  l«  Cl  II  II 

10 

9 

Rule  to  find  teeth  required  in 

8 

12 

II  <1  II  Cl  II 

11 

22 

“  Rules  applying  to . 

8 

10 

“  “  the  same  counts  . 

9 

8 

for  draft . 

15 

26 

II  cc  cc  cc  <*  c* 

10 

8 

“  Spiral . 

7 

32 

Cl  Cl  CC  CC  Cl  Cl 

11 

20 

“  Sprocket  . 

7 

31 

Fraction,  Complex . 

2 

3 

“  Spur . 

7 

30 

Improper . 

2 

3 

“  Star  . 

7 

32 

Proper  . 

2 

2 

“  Trains  of . 

7 

29 

Simple . 

2 

3 

Greatest  common  divisor . 

1 

29 

Terms  of  a . 

2 

2 

Groove  pulleys . 

7 

16 

to  a  decimal,  Reducing  a 

2 

29 

Guide  pulley . 

7 

17 

Value  of  a . 

2 

2 

Gunter’s  chain . 

5 

6 

Fractions . 

2 

1 

II 

Addition  of . 

2 

9 

Common  denomina- 

Hanger,  Wall-box . 

7 

9 

tor  of  . 

2 

8 

Hangers,  Distance  between  .  .  . 

7 

7 

Division  of . 

2 

15 

Shaft . 

7 

5 

Multiplication  of  ...  . 

2 

13 

Types  of . 

7 

7 

Reduction  of . 

2 

4 

Hanks  of  filling,  Finding . 

12 

12 

Roots  of . 

4 

15 

CC  CC  Cl  II 

12 

20 

Subtraction  of . 

2 

11 

“  '!  warp  yarn.  Finding  .  .  . 

12 

18 

Freehand  drawings  .  .  .  • . 

91 

2 

“  Finding  num- 

Friction  clutches . 

7 

2 

ber  of  .  .  . 

12 

7 

II  II 

7 

5 

Harness . 

12 

i 

“  couplings . 

7 

5 

c  c 

13 

2 

Frictional  contact  . . 

7 

9 

calculations . 

12 

3 

Front  rolls . 

15 

6 

II  4  1 

13 

4 

Frustum  of  a  pyramid  or  cone  .  . 

6 

14 

“  draft . 

12 

3 

Full  line . 

91 

22 

Cl  cc 

13 

4 

Fundamental  principles  of  drawing 

91 

3 

Head,  or  main,  shaft . 

7 

2 

XVI 


INDEX 


Sec. 

Page 

Sec.  Page 

Heavy  full  line . 

91 

22 

Line,  Light  full . 

.  91 

22 

Heddles . 

12 

3 

“  shafts . 

.  7 

2 

Heel  of  a  cam . 

7 

36 

Linear  or  long  measure . 

.  5 

5 

Heptagon . 

6 

7 

Linen,  jute,  and  ramie . 

.  11 

16 

Hexagon . 

6 

7 

Lines . 

6 

1 

Horizontal  line,  A . 

6 

1 

“  Broken-and-dotted  .  .  .  . 

.  91 

22 

Horsepower  transmitted  by  belts  . 

8 

7 

Dimension  . 

.  91 

27 

Hundreds . 

1 

3 

Dot-and-dash . 

.  91 

22 

f 

Extension . 

.  91 

27 

“  Limiting . 

.  91 

27 

Idle  gear . 

7 

35 

“  Parallel . 

6 

1 

“  pulley . 

7 

17 

“  Shade  . 

.  91 

22 

Idler . 

7 

17 

used  on  drawings . 

.  91 

22 

Improper  fraction . 

2 

3 

Liquid  measure . 

5 

4 

Index  of  the  root . 

4 

3 

Listing . 

.  13 

7 

Integer . 

1 

2 

Local  value . 

1 

3 

Integral  number . 

1 

2 

Long  division . 

1 

23 

Interest . 

3 

7 

“  or  linear  measure . 

.  5 

5 

Intermediate  gear . 

7 

35 

Longitudinal  sections . 

.  91 

15 

Inverse  proportion . 

3 

13 

Loom  beams . 

9 

13 

3 

16 

4  4  4  4 

.  11 

30 

ratio . 

3 

11 

“  Weight  of  cloth  from  .  .  . 

.  13 

24 

Inverted  pillow-block . 

7 

9 

Loose  and  tight  pulleys . 

.  7 

17 

plan . 

91 

7 

“  couplings . 

.  7 

2 

Involution . 

4 

1 

Lumber,  Mensuration  of  .... 

6 

16 

Irrational  factors . 

4 

8 

Isosceles  triangle . 

6 

3 

M 

J 

Machine  elements . 

.  7 

1 

Main,  or  head,  shaft . 

7 

2 

Jute,  linen,  and  ramie . 

11 

16 

Mangle  gears . 

7 

32 

K 

Measure,  Angular . 

.  5 

7 

Key  seats . 

7 

2 

Apothecaries’  fluid  .  . 

.  5 

4 

Keys . 

7 

2 

Board . 

6 

16 

Keyways . 

7 

2 

Circular . 

.  5 

7 

Cubic . 

.  5 

6 

L 

“  Dry . 

.  5 

5 

Law  of  value  expressed  by  figures 

1 

4 

Linear  or  long . 

.  5 

5 

Lay . 

13 

2 

Liquid . 

.  5 

4 

Least  common  denominator  of 

of  money . 

.  5 

2 

fractions  .  .  . 

2 

8 

“  time . 

.  5 

7 

multiple  .... 

1 

31 

Square . 

.  5 

6 

Leather  belts . 

7 

10 

Surface . 

.  5 

6 

Left-handed  threads . 

91 

30 

Surveyors'  . 

.  5 

5 

Length,  Measures  of . 

5 

5 

Measures . 

.  5 

2 

“  of  belts . 

8 

6 

Miscellaneous  ... 

.  5 

8 

Letters,  Reference . 

91 

33 

of  area . 

.  5 

6 

Lever,  Pressure  exerted  by  a  .  .  . 

8 

16 

“  length . 

.  5 

5 

Levers  . 

7 

37 

“  quantity . 

.  5 

4 

“  Rules  applying  to . 

8 

16 

“  solidity  or  volume  . 

.  5 

6 

Light  full  line . 

91 

22 

“  weight . 

.  5 

3 

“  surface . 

91 

23 

Mechanical  calculations . 

.  8 

1 

Like  numbers . 

1 

1 

definitions . 

.  7 

1 

Limiting  lines . 

91 

27 

drawings  . 

.  91 

3 

Line,  Broken  . 

91 

22 

Mensuration . 

.  6 

1 

“  Dotted . 

91 

22 

of  lumber . 

6 

16 

“  Heavy  full . 

91 

22 

“  solids . 

6 

10 

INDEX 


XVII 


Sec.  Page 

Sec.  Page 

Mensuration  of  surfaces . 

6 

1 

Number  of  the  reed . 

13 

6 

Metallic  rolls  . 

15 

20 

“  “  warp,  Average  .  . 

10 

23 

Method  of  numbering-  ply  yarns  .  . 

9 

7 

“  yarns.  Average  .  .  .  . 

9 

15 

<«  44  14  44  14 

10 

7 

44  44  44  44 

11 

33 

44  44  44  44  44 

11 

20 

Power  of  a . 

4 

1 

Methods  of  finding  draft . 

15 

7 

Prime . 

1 

27 

Metric  system  of  numbering  yarns 

9 

21 

Reciprocal  of  a . 

S 

11 

44  44  44  44  44 

10 

25 

Root  of  a . 

4 

1 

44  44  44  44  44 

11 

38 

Square  of  a . 

4 

1 

Minuend . 

1 

12 

Unit  of  a . 

1 

1 

Minus  sign . 

1 

12 

Numbering  ply  yarns . 

9 

7 

Miscellaneous  measures . 

5 

8 

44  44  44 

10 

7 

Miter  gears . 

7 

31 

44  44  44 

11 

20 

Mixed  number . 

2 

3 

single  worsted  yarns  . 

10 

1 

Money,  Measure  of . 

5 

2 

yarns  . 

10 

13 

“  Reduction  of  United  States 

6 

11 

system . 

9 

1 

“  United  States . 

5 

2 

“  Yarn . 

11 

2 

Mortise  gears . 

7 

28 

woolen  yarns,  Cut  sys- 

Mover,  First,  or  prime . 

7 

2 

tem  of . 

10 

14 

Movers,  Second . 

7 

2 

woolen  yarns.  Cut  sys- 

Moving  parts  shpwn  in  two  or 

tem  of . 

11 

9 

more  positions . 

91 

33 

woolen  yarns.  Run 

Muff  couplings  . 

7 

2 

system  of . 

10 

13 

Mule  pulley  stand . 

7 

24 

woolen  yarns,  Run 

Multiple,  Common  . 

1 

31 

system  of . 

11 

7 

threaded  screw . 

7 

35 

yarns,  Metric  system 

Multiplicand . 

i 

16 

of . 

9 

21 

Multiplication . 

i 

16 

“  Metric  system 

of  compound  quan- 

of . 

10 

25 

tities . 

5 

14 

“  Metric  system 

“  decimals  .... 

2 

23 

of . 

11 

38 

“  fractions  .  .  .  . 

2 

13 

Numbers,  Borrowing . 

1 

13 

Proof  of . 

1 

20 

Compound . 

5 

1 

Rule  for . 

1 

19 

Denominate . 

5 

1 

Sign  of . 

1 

16 

“  Like . 

1 

1 

“  table . 

1 

17 

Simple . 

5 

1 

Multiplier . 

1 

16 

To  add  compound  .  .  . 

5 

13 

N 

Unlike . 

1 

1 

Numeration . 

1 

5 

Naught  . 

1 

2 

and  notation . 

1 

2 

Non-parallel  shafts . 

7 

22 

Numerator 

2 

1 

“  positive  cam . 

7 

36 

O 

Notation . 

i 

5 

and  numeration . 

i 

2 

Objects,  Conventional  methods  of 

Arabic . 

i 

2 

representing . 

91 

28 

Number,  A  . 

i 

1 

“  of  drafting . 

15 

2 

Composite . 

i 

27 

Representation  of  ...  . 

91 

1 

Cube  of  a . 

4 

1 

Obtuse  angle  . 

6 

2 

Even . 

1 

28 

Octagon . 

6 

7 

Exponent  of  a . 

4 

1 

Odd  number . 

1 

28 

Integral . 

1 

2 

Open  belt . 

7 

20 

Mixed . 

2 

3 

Ounce  cloth . 

12 

2 

"  Odd . 

1 

28 

of  ends  in  a  warp  .  .  .  . 

12 

5 

P 

“  “  “  the  warp  .  .  . 

13 

22 

Parallel  lines . 

6 

1 

“  hanks  of  warp  yarn  . 

12 

7 

Parallelogram . 

6 

4 

XV111 


INDEX 


Sec. 

Page 

Sec. 

Page 

Parallelopipedon  . 

6 

10 

Polygon,  Area  of  a  regular  .... 

6 

7 

Parenthesis . 

19 

Regular . 

6 

7 

Part,  Significant . 

4 

16 

Polygons  . 

6 

7 

Partial  products . 

1 

19 

Positive-motion  cam  . 

7 

36 

“  sections . 

91 

14 

Post  hangers  . 

7 

7 

Parts  shown  in  two  or  more  posi- 

Pound,  Yards  of  cotton  cloth  per  . 

12 

21 

tions,  Moving . 

91 

33 

Power  of  a  number . 

4 

1 

Pattern  of  the  warp . 

9 

17 

“  transmission . 

7 

9 

it  <  <  <  (  1 1 

10 

22 

of . 

8 

1 

“  “  “  “ 

11 

35 

Powers,  Perfect . 

4 

8 

Patterns,  Fancy . 

13 

14 

Prime  factor . 

1 

27 

Pentagon  . 

6 

7 

“  number . 

1 

27 

Per  cent.,  Definition  of  . 

3 

1 

“  or  first  mover . 

7 

2 

"  “  Sign  of . 

3 

2 

Principal . 

3 

7 

Percentage  . 

3 

1 

Principles  of  drawing,  Funda- 

base,  and  rate . 

3 

4 

mental . 

91 

3 

Rule  to  find . 

3 

4 

“  percentages  .  .  .  . 

3 

1 

Perfect  powers  . 

4 

8 

Prism,  The . 

6 

10 

Permanent  couplings . 

7 

2 

Product  . 

1 

16 

Perpendicular  line,  A . 

6 

1 

Production  and  draft,  Calculating 

15 

22 

Perspective  views . 

91 

18 

Products,  Partial  . 

1 

19 

Pick . 

13 

2 

Proof  of  addition  . 

1 

11 

“  cloth  . 

12 

2 

“  “  division . 

1 

25 

Picks,  Definition  of . 

12 

2 

“  “  multiplication . 

1 

20 

Pillow-block . 

7 

9 

“  “  square  root  . 

4 

9 

Pinion,  A . 

7 

34 

“  “  subtraction . 

1 

14 

Pitch  circle  of  a  gear . 

7 

28 

"  “  weight  and  counts  .  .  .  . 

13 

22 

“  of  a  gear . 

7 

29 

Proper  fraction . 

2 

2 

“  44  44  screw . 

7 

35 

Proportion . 

3 

12 

Plan,  Inverted . 

91 

7 

Inverse . 

3 

16 

“  Top . 

91 

7 

Unit  method  of  .... 

3 

18 

Plane  figure,  A . 

6 

2 

Pulley,  Binder . 

7 

17 

“  surface,  A . 

6 

2 

“  bushings . 

7 

15 

Plans . 

91 

7 

“  Crown-faced . 

7 

14 

Plumb-line . 

6 

1 

Flat . 

7 

14 

Ply  yarns  . 

9 

7 

“  Guide  . 

7 

17 

10 

7 

Idle  . 

7 

17 

10 

17 

“  Sheave  . 

7 

16 

11 

20 

“  stand,  Mule . 

7 

24 

"  Calculation  of  cost  of  .  . 

11 

25 

“  Straight-faced . 

7 

14 

“  Calculations  of . 

9 

8 

Pulleys . 

7 

13 

44  II  II  II 

10 

8 

“  Flange  and  groove  .... 

7 

16 

cotton 

11 

20 

“  Ratios  of  speeds  to  diam- 

“  composed  of  more  than 

eters  of  . 

8 

5 

two  threads . 

9 

9 

“  Solid . 

7 

15 

"  composed  of  more  than 

“  Speed  of . 

7 

22 

two  threads . 

10 

10 

“  Speeds  of . 

8 

2 

"  composed  of  more  than 

“  Split . 

7 

15 

two  threads . 

11 

23 

“  Step . 

7 

18 

Method  of  numbering  .  . 

9 

7 

Surface  velocity  of  rotating 

8 

5 

II  14  It  II  II 

10 

7 

“  Tight  and  loose . 

7 

17 

14  II  II  II  II 

11 

20 

Pyramid  and  cone . 

6 

13 

"  of  different  materials  .  . 

11 

27 

“  “  spun  silk . 

11 

26 

Q 

“  “  Woolen  and  worsted  .  . 

11 

26 

Quadrilaterals . 

6 

4 

Points,  Gauge . 

10 

6 

Quantities,  Addition  of  compound 

5 

12 

INDEX 


xix 


Sec. 

Page 

Sec.  Page 

Quantities,  Division  of  compound 

5 

16 

Repeat  of  colors . 

9 

17 

Multiplication  of  com- 

Representation  of  objects . 

91 

1 

pound  . 

5 

14 

Rhomboid . 

6 

4 

Reduction  of  com- 

Rhombus . 

6 

4 

pound  . 

5 

9 

Ribs . 

12 

4 

Subtraction  of  com- 

“ 

13 

5 

pound  . 

5 

13 

Right  angle . 

6 

2 

Quantity,  Measures  of . 

5 

4 

“  handed  threads . 

91 

30 

Quarter-turn  belt 

7 

22 

triangle . 

6 

3 

Quotient . 

1 

22 

Rolls  and  gears  on  draft,  Effect  of 

15 

14 

Back . 

15 

6 

R 

“  Calculating  draft  for  com- 

Rack,  A . 

7 

34 

mon . 

15 

12 

Radical  sign . 

4 

3 

“  Delivery . 

15 

6 

Radius  of  a  circle . 

6 

8 

“  Drafting  with  common  .  .  . 

15 

5 

Ramie,  jute,  and  linen . 

11 

16 

“  Front . 

15 

6 

Ratchet  gear . 

7 

31 

“  Gearing  of . 

15 

9 

Rate,  base,  and  percentage  .... 

3 

4 

"  Metallic . 

15 

20 

“  of  interest . 

7 

“  on  draft,  Effect  of  diameters 

“  per  cent . 

3 

3 

of . 

15 

12 

Ratio . 

10 

“  Top . 

15 

6 

Rational  factors . 

4 

8 

Root,  Cube . 

4 

3 

Ratios  of  speeds  to  diameters  of 

“  “ 

4 

10 

pulleys . 

8 

5 

“  Index  of  the  . 

4 

3- 

Raw  silk . 

11 

14 

“  of  a  number . 

4 

1 

Reading  textile  drawings . 

91 

1 

“  Square  . 

4 

3 

Reciprocal  of  a  number . 

3 

11 

Roots,  Extracting  square  and  cube 

4 

16 

ratio . 

8 

11 

“  of  fractions . 

4 

15 

Reciprocating  screw . 

7 

35 

Rope  drives . 

7 

24 

Rectangle . 

6 

4 

“  transmission . 

7 

24 

Reducing  a  decimal  to  a  fraction, 

4<  <1 

8 

9- 

Rule  for . 

2 

30 

Rotation,  Direction  of  pulley  .  .  . 

7 

20 

fraction  to  a  decimal 

2 

29 

Roving  and  yarn,  Sizing . 

9 

3 

“  its  lowest 

cotton  .  . 

11 

4 

terms  . 

2 

5 

“  woolen 

10 

15- 

Reduction . 

5 

1 

44  44  44  44  44 

11 

9- 

of  compound  quantities 

5 

9 

"  Definition  of . 

9 

3 

“  United  States  money 

5 

11 

“  Sizing  worsted . 

10 

6 

Reductions  of  fractions . 

2 

4 

“  “  “ 

11 

13 

Reeds  .  . 

12 

4 

Rubber  belts . 

7 

10- 

“  and  ends  per  dent . 

13 

24 

Rule  for  addition . 

1 

10 

“  calculations . 

13 

5 

“  for  addition  of  compound 

“  drafts,  Irregular . 

12 

12 

numbers . 

5 

13 

“  Number  of  the . 

13 

6 

“  for  addition  of  decimals  .  .  . 

2 

21 

“  table . 

12 

17 

"  for  addition  of  fractions  .  .  . 

2 

10 

“  Width  at . 

12 

10 

“  for  allowing  for  shrinkage 

41  in . 

13 

24 

of  cloth  or  yarn . 

13 

11 

Reeds  . 

12 

4 

“  for  cancelation . 

1 

35 

Reel,  Wrap . 

9 

5 

“  for  division . 

1 

25 

<1  4  4> 

10 

4 

“  for  division  of  compound 

4  4  4  1 

11 

5 

numbers . 

5 

16 

Reference  letters  . 

91 

33 

for  division  of  decimals  .  .  . 

2 

27 

Regular  polygon . 

C 

7 

“  for  division  of  fraction  by 

Relative  value . 

1 

3 

whole  number . 

2 

15 

Remainder . \ 

1 

12 

“  for  division  of  whole  number 

“ 

1 

22 

or  fraction  by  fraction  .  .  . 

2 

15- 

XX 


INDEX 


Sec.  Page  Sec.  Page 


for  extracting  other  roots  than 

Rule  to  find  base  when  percentage 

square  and  cube . 

4 

19 

and  rate  are  known  .... 

3 

5 

for  extracting  roots  of  frac- 

to  findcircumferenceof  a 

tions . 

4 

16 

circle  . 

6 

8 

for  filling  required  to  preserve 

“  to  find  common  divisor  .  .  . 

1 

29 

the  weight  of  cloth . 

12 

25 

to  find  cost  of  two-ply  yarns 

for  finding  cube  root . 

4 

14 

when  threads  are  of  differ- 

for  finding  square  root  .  .  .  . 

4 

10 

ent  values . 

11 

26 

for  length  of  belts . 

8 

6 

to  find  count  of  one  system 

for  multiplication . 

1 

19 

equivalent  to  that  of  an- 

for  multiplication  of  com- 

other . 

11 

18 

pound  quantities . 

5 

15 

to  find  counts  in  ply  yarns  .  . 

9 

10 

for  multiplication  of  decimals 

2 

24 

to  find  counts  of  a  roving  .  . 

10 

6 

for  multiplication  of  fractions 

2 

14 

to  find  counts  of  a  roving  .  . 

11 

13 

for  raising  a  number  or  deci- 

to  find  counts  of  a  yarn  to 

mal  to  any  power . 

4 

3 

be  folded  with  another  .  .  . 

11 

24 

for  reducing  decimal  to  frac- 

to  find  counts  of  a  yarn  when 

tion . 

.  2 

30 

length  and  weight  are  given 

9 

2 

for  reducing  fraction  to  deci- 

to  find  counts  of  a  yarn  when 

mal . 

2 

29 

length  and  weight  are  givbn 

10 

2 

for  reducing  fraction  to  low- 

to  find  counts  of  a  yarn  when 

est  terms . 

2 

5 

length  and  weight  are  given 

11 

3 

for  reducing  fractions  to  frac- 

to  find  counts  of  filling  to  pre- 

tions  with  common  denom- 

serve  weight . 

12 

23 

inator . 

2 

9 

to  find  counts  of  yarn  on  a 

for  reducing  inches  to  deci- 

beam  containing  only  one 

mal  part  of  foot . 

2 

30 

size  of  yarn . 

11 

31 

for  reduction  of  compound 

to  find  counts  of  yarn  on  a 

quantities  to  lower  denomi- 

beam  that  contains  one 

nations . 

5 

10 

size  of  yarn . 

9 

13 

for  reduction  of  a  quantity  to 

“  to  find  dents  per  inch  in  a 

higher  denominations  .  .  . 

5 

11 

reed  . 

12 

10 

for  rope  transmission  .... 

8 

9 

“  to  find  diameter  of  circle  with 

for  speed  of  worm-gears  .  .  . 

8 

12 

given  circumference  .... 

6 

9 

for  subtraction . 

1 

14 

“  to  find  diameter  of  a  driven 

for  subtraction  of  compound 

pulley . 

8 

3 

quantities . 

5 

14 

“  to  find  diameter  of  a  driving 

for  subtraction  of  fractions  . 

2 

12 

pulley . 

8 

4 

for  subtraction  of  decimals  . 

2 

22 

“  to  find  diameter  of  counter- 

to  find  a  constant . 

8 

15 

shafts . 

8 

1 

to  find  area  of  circle . 

6 

9 

“  to  find  diameter  of  gear-blank 

8 

12 

to  find  area  of  parallelogram 

6 

5 

“  to  find  diameter  of  line  shafts 

8 

1 

to  find  area  of  regular  polygon 

6 

7 

to  find  greatest  common  di- 

to  find  area  of  trapezium  .  . 

6 

6 

visor . 

1 

30 

to  find  area  of  trapezoid  .  .  . 

6 

5 

“  to  find  least  common  multiple 

1 

32 

to  find  area  of  triangle  .... 

6 

3 

“  to  find  percentage . 

3 

4 

to  find  average  counts  of  ends 

“  to  find  prime  factors . 

1 

28 

on  a  beam . 

11 

33 

to  find  rate  when  percentage 

to  find  average  counts  of  yarn 

and  base  are  known  .... 

3 

5 

in  a  piece  of  cloth . 

12 

21 

to  find  size  of  cold-rolled 

to  find  average  number  of 

shafting . 

8 

1 

warp  yarn . 

9 

15 

to  find  the  diameter  of  yarns  . 

11 

30 

to  find  average  number  of 

“  to  find  the  draft  between  two 

yarn  in  a  cloth . 

12 

24 

pair  of  rolls . 

15 

14 

to  find  average  size  of  warp 

to  find  the  draft  constant  of  a 

yarn . 

10 

23 

machine . 

15 

28 

INDEX 


xxi 


Sec.  Page 

Sec. 

Page 

Rule  to  find  the  draft  gear . 

15 

29 

Rule  to  find  the  number  of  feet  of 

“  to  find  the  draft  of  a  machine 

15 

31 

lumber  in  1-inch  boards  .  . 

6 

16 

to  find  the  hank  of  a  roving 

15 

32 

(  « 

to  find  the  number  of  heddles 

to  find  the  hanks  of  filling 

required  on  any  harness  .  . 

12 

3 

contained  in  a  cloth  .... 

12 

12 

4  4 

to  find  the  number  of  heddles 

“  to  find  the  hanks  of  warp  yarn 

required  on  each  harness  . 

13 

4 

contained  in  a  cloth  .... 

12 

9 

(  < 

to  find  the  number  of  revolu- 

“  to  find  the  horsepower  trans- 

tions  per  minute  of  a  driven 

mitted  by  a  belt . 

8 

7 

pulley . 

8 

3 

“  to  find  the  length  of  a  square 

6 

10 

il 

to  find  the  number  of  teeth  re- 

to  find  the  length  of  an  in- 

quired  in  gears . 

8 

12 

scribed  square . 

6 

9 

4 1 

to  find  the  number  of  yards 

“  to  find  the  length  of  a  warp  . 

10 

20 

per  pound  in  a  sample  of 

to  find  the  length  of  a  warp  . 

11 

32 

cloth . 

12 

14 

to  find  the  length  of  warp  on 

4 4 

to  find  the  number  of  yarn  in 

a  beam . 

9 

14 

any  system  corresponding 

“  to  find  the  length  of  warp  re- 

to  number  in  metric  sys- 

quired  for  a  certain  length 

tern . 

9 

22 

of  cloth . 

13 

10 

4 1 

to  find  the  number  of  yarn  in 

“  to  find  the  length  of  warp  that 

any  system  corresponding 

can  be  placed  on  a  beam  . 

9 

14 

to  number  in  metric  sys- 

“  to  find  the  length  of  waip  that 

tern . 

10 

25 

can  be  placed  on  a  beam  • 

10 

20 

“ 

to  find  the  number  of  yarn  in 

“  to  find  the  length  of  warp  that 

any  system  corresponding 

can  be#  placed  on  a  beam 

11 

32 

to  number  in  metric  sys- 

“  to  find  the  length  of  yarn  .  .  . 

9 

2 

tem . 

11 

38 

“  to  find  the  length  of  yarn  .  .  . 

10 

3 

4  4 

to  find  the  number  of  yarn  in 

“  to  find  the  length  of  yarn  .  .  . 

11 

3 

metric  system  correspond- 

“  to  find  the  number  of  counts 

ing  to  number  of  yarn  in 

' 

in  a  ply  yarn . 

10 

10 

another  system . 

9 

22 

“  to  find  the  number  of  ends  of 

4  4 

to  find  the  number  of  yarn  in 

each  color  of  yarn  on  a 

metric  system  correspond- 

beam  . 

10 

21 

ing  to  number  of  yarn  in 

“  to  find  the  number  of  ends  of 

another  system . 

10 

26 

each  color  of  yarn  on  a 

to  find  the  number  of  yarn  in 

beam . 

9 

16 

metric  system  correspond- 

“  to  find  the  number  of  ends  of 

ing  to  number  of  yarn  in 

each  color  of  yarn  on  a 

another  system . 

11 

39 

beam . 

11 

36 

44 

to  find  the  original  size  of  a 

“  to  find  the  number  of  ends  of 

yarn . 

13 

13 

each  count  or  color  in  a 

44 

to  find  the  percentage  of  warp 

cloth  . 

13 

14 

contraction  during  weaving 

12 

9 

“  to  find  the  number  of  ends  of 

“ 

to  find  the  pressure  exerted 

each  kind  in  the  body  of  the 

by  a  lever . 

8 

16 

cloth . 

12 

7 

“ 

to  find  the  reed  to  use  for  an 

“  to  find  the  number  of  ends  on 

unevenly  reeded  cloth  .  .  . 

13 

8 

a  beam . 

10 

19 

11 

to  find  the  reed  when  the  ends 

“  to  find  the  number  of  ends  on 

per  inch  in  the  loom  are 

a  beam . . 

9 

14 

known . 

13 

15 

“  to  find  the  number  of  ends  on 

4  4 

to  find  the  reed  when  the  width 

a  beam . 

11 

31 

in  the  loom  is  known  .  .  . 

13 

8 

“  to  find  the  number  of  ends  per 

“ 

to  find  the  required  weight  of 

inch  in  the  reed . 

13 

15 

each  thread  folded  to  pro- 

“  to  find  the  number  of  feet  of 

duce  a  ply  yarn . 

11 

24 

lumber  in  joists,  beams. 

4  4 

to  find  the  required  width  of 

etc . 

6 

16 

belt . 

8 

8 

XXII 


INDEX 


Sec. 

Page 

Sec.  Page 

Rule  to  find  the  resultant  count  of 

Rule 

to  find  the  volume  of  the 

ply  yarns . 

ii 

23 

frustum  of  a  pyramid  or 

4  4 

to  find  the  resultant  counts 

cone . 

6 

14 

when  different  counts  are 

44 

to  find  the  weight  of  a  sliver 

folded . 

9 

10 

or  roving  produced  by  a 

4  4 

to  find  the  resultant  counts 

machine . 

15 

31 

when  different  counts  are 

to  find  the  weight  of  a  warp 

10 

19 

folded . 

10 

10 

44 

to  find  the  weight  of  filling  in 

<  4 

to  find  the  resultant  counts 

a  yard  of  cloth . 

13 

17 

when  different  counts  are 

4  4 

to  find  the  weight  of  single 

folded . . . 

11 

24 

yarns  required  to  produce  a 

4  4 

to  find  the  resultant  counts 

ply  yarn . 

9 

11 

when  two  threads  are 

4  4 

to  find  the  weight  of  the  warp 

folded . 

9 

9 

in  a  yard  of  cloth . 

13 

17 

(4 

to  find  the  resultant  counts 

4  4 

to  find  the  weight  of  threads 

when  two  yarns  are  folded 

10 

9 

required  to  produce  a  given 

4  4 

to  find  the  revolutions  of  the 

count . 

10 

11 

driving:  shaft . 

8 

3 

4  4 

to  find  the  weight  of  yarn  on 

4  4 

to  find  the  size  of  a  given  yarn 

12 

22 

a  beam  . 

9 

14 

4  4 

to  find  the  size  of  a  given 

4  4 

to  find  the  weight  of  yarn  on 

yarn . 

13 

16 

a  beam  . 

11 

32 

4  4 

to  find  the  size  of  a  woolen 

41 

to  find  the  weight  of  yarn 

roving:  or  yarn . 

10 

17 

when  length  and  counts  are 

4  4 

to  find  the  size  of  a  woolen 

given . 

9 

2 

roving:  or  yarn . 

11 

10 

4  4 

to  find  the  weight  of  yarn 

4  4 

to  find  the  size  of  yarn  on 

when  the  length  and  counts 

jack  spool  or  beam  con- 

are  known . 

11 

3 

taining  only  one  size  of 

4  4 

to  find  the  weight  per  yard  of 

yarn . 

10 

18 

finished  cloth . 

13 

17 

4  4 

to  find  the  speed  of  a  driven 

44 

to  find  the  width  in  the  loom 

g:ear . 

8 

10 

inside  selvages . 

13 

15 

4  4 

to  find  the  speed  of  a  driven 

4  4 

to  find  the  width  occupied  by 

pulley . 

8 

4 

the  warp  yarn  in  the  reed 

12 

11 

4  4 

to  find  the  standard  breaking 

4  4 

to  find  the  width  of  cloth  in 

weight  of  warp  yarns  .  .  . 

9 

19 

the  reed . 

13 

7 

4  4 

to  find  the  surface  area  of  a 

4  4 

to  find  the  weight  of  a  yarn  . 

10 

2 

cylinder . 

6 

12 

4  4 

to  find  what  counts  must  be 

4  4 

to  find  the  surface  area  of  a 

folded  to  produce  a  given 

prism  . 

6 

11 

count . 

10 

11 

4  4 

to  find  the  surface  area  of  a 

4  4 

to  find  what  counts  must  be 

sphere  . 

6 

15 

folded  to  produce  given 

4  4 

to  find  the  surface  velocity  of 

counts  . 

9 

11 

rotating  pulleys . 

8 

5 

4  4 

to  reduce  a  fraction  to  a 

4  4 

to  find  the  total  ends  in  a 

given  denominator  .... 

2 

6 

cloth . 

12 

6 

4  4 

to  reduce  a  mixed  number  to 

4  4 

to  find  the  total  number  of 

an  improper  fraction  .  .  . 

2 

6 

ends  in  a  warp . 

13 

8 

“ 

to  reduce  an  improper  frac- 

4  4 

to  find  the  twist  to  be  inserted 

tion  to  a  whole  or  mixed 

in  any  counts  of  yarn  .  .  . 

9 

18 

number . 

2 

7 

4  4 

to  find  the  volume  of  a  cone 

4  4 

to  reduce  a  whole  number  to 

or  pyramid  . 

6 

14 

an  improper  fraction  .  .  . 

2 

6 

41 

to  find  the  volume  of  a  cylin- 

Rules  applying  to  belts . 

8 

6 

der . 

6 

13 

44 

“  gears  . 

8 

10 

to  find  the  volume  of  a  prism 

6 

11 

4  4 

“  levers  . 

8 

16 

to  find  the  volume  of  a 

44 

shafting  .... 

8 

1 

sphere  . . 

6 

15 

44 

“  speeds  . 

8 

2 

INDEX  xxiii 


Sec.  Page 

Sec.  Page 

Rules  for  draft  gears . 

15 

26 

Shrinkage  in  finishing . 

13 

11 

“  “  proportion  . 

3 

13 

Side  views . 

91 

6 

“  “  the  reduction  of  United 

Significant  figures . 

4 

16 

States  money . 

5 

11 

part . 

4 

16 

“  pertaining  to  the  transmis- 

Sign,  Minus . 

1 

12 

sion  of  power . 

8 

1 

“  of  addition . 

1 

6 

“  to  find  size  of  turned  shaft- 

“  division . 

1 

22 

ing  for  given  horsepower  . 

8 

2 

equality . 

1 

6 

used  in  obtaining  particu- 

multiplication  .... 

1 

16 

lars  of  cloth  samples  .  . 

12 

24 

“  “  per  cent . 

3 

2 

Run  scale . 

11 

9 

“  “  subtraction . 

1 

12 

“  system  of  numbering  woolen 

Radical . 

4 

3 

yarns  . 

10 

13 

Signs . 

1 

2 

“  numbering  woolen 

“  of  aggregation  . 

3 

19 

yarns  . 

11 

7 

Silk,  Ply  yarns  of  spun  .... 

11 

26 

Runs,  Table  of . 

11 

11 

“  Varieties  of  . 

11 

14 

“  yarns  . 

11 

14 

S 

Simple  fraction . 

2 

3 

Samples,  Figuring  particulars 

“  numbers . 

5 

1 

from  cloth . 

12 

16 

“  value . 

1 

2 

Figuring  particulars 

Single  and  double  belts  .... 

7 

11 

from  cloth . 

13 

21 

“  cotton  yarns . 

n 

3 

Rules  used  in  obtaining 

“  spun  silk . 

ii 

14 

particulars  of  cloth  .  . 

12 

24 

“  width  cloth . 

13 

3 

Scale,  Cut . 

11 

9 

“  worsted  yarns  .... 

10 

1 

“  Run . 

11 

9 

“  yarns  . 

9 

1 

Scalene  triangle . 

6 

3 

11  n 

11 

2 

Scales  for  drawing . 

91 

26 

“  Calculations  of  . 

9 

2 

Screw,  Differential . 

7 

35 

“  Systems  of  number- 

“  Multiple-threaded . 

7 

35 

ing . 

10 

13 

“  Pitch  of  a . 

7 

35 

Size,  Allowance  for  weight  of  . 

12 

19 

“  Reciprocating . 

7 

35 

“  of  gears  on  draft.  Effect 

of 

15 

13 

Screws . 

7 

35 

Sizing  cotton  roving  and  yarn 

11 

4 

Seats,  Key . 

7 

2 

“  gear-blanks . 

8 

12 

Second  movers . 

7 

2 

“  roving  and  yarn  .... 

9 

3 

Section  beams . 

9 

13 

“  woolen  roving  and  yarn 

10 

15 

Sectional  views . 

91 

11 

<<  it  a  a  n 

11 

9 

Sections,  Longitudinal . 

91 

15 

“  worsted  roving . 

10 

6 

Partial . 

91 

14 

“  “  “ 

11 

13 

Selvage  ends  . 

12 

6 

“  yarns  . 

10 

3 

Selvages . 

13 

7 

(4  44  44 

11 

12 

Semicircle . 

6 

8 

Sley . 

12 

16 

Semi-circumference . 

6 

8 

“  Average . 

12 

23 

Shade  lines  . 

91 

22 

“  Definition  of . 

12 

2 

Shaft,  A . 

7 

1 

Solid,  A . 

6 

10 

“  couplings . 

7 

2 

“  pulleys . 

7 

15 

“  hangers  . 

7 

5 

Solidity,  or  volume.  Measures  of 

5 

6 

“  Main  or  head . 

7 

2 

Solids,  Mensuration  of  .... 

6 

10 

Shafting . 

7 

1 

Speed  cones . 

7 

18 

Rules  applying  to  .... 

8 

1 

“  Effect  of  countershafts 

on 

8 

4 

Shafts,  Line . 

7 

2 

“  of  pulleys  ...... 

7 

22 

“  Non-parallel . 

7 

22 

“  “  worm-gears . 

8 

12 

Sheave  pulley . 

7 

16 

Speeds  of  pulleys . 

8 

2 

Shed . 

13 

2 

“  Ratios  of,  to  diameters 

of 

Shipper,  Belt . .  .  .  .  . 

7 

18 

pulleys  . 

8 

5 

Short  division . 

1 

23 

“  Rules  applying  to  .  .  . 

8 

2 

XXIV 


INDEX 


Sec.  Page 

Sec. 

Page 

Sphere  . 

6 

15 

Table  of  liquid  measure . 

5 

4 

Spiral  gears . .  .  . 

J 

32 

“  “  mill  weights . 

5 

3 

Split . 

13 

6 

runs . 

11 

11 

“  pulleys . 

7 

15 

“  “  sizing  for  cotton  roving 

Sprocket  gears  . 

7 

31 

or  yarn  . 

11 

7 

wheels . 

7 

31 

“  “  sizing  for  worsted  yarns  . 

11 

12 

Spun  silk,  Ply  yarns  of . 

ii 

26 

“  “  square  measure . 

5 

6 

“  Single . 

ii 

14 

“  squares  and  cubes  .  .  .  . 

4 

17 

Spur  gears . 

7 

30 

surveyors’  measure  .  .  . 

5 

5 

Square  . 

6 

4 

“  time  measure . 

5 

7 

<  < 

12 

2 

“  “  Troy  weight . 

5 

3 

“  Definition  of  a . 

5 

6 

“  “  United  States  money .  .  . 

5 

2 

“  roots.  Extracting . 

4 

16 

“  Reed . 

12 

17 

measure . 

5 

6 

Teeth . 

7 

28 

“  of  a  number . 

4 

1 

“  required  in  gears,  Rule  to 

“  root  . 

4 

3 

find . 

8 

12 

“  Proof  of . 

4 

9 

Tens . 

1 

3 

Squares  and  cubes . 

4 

17 

Terms  of  a  fraction . 

2 

2 

Star  gears . 

7 

32 

“  “  “  ratio . 

3 

10 

Step  pulleys . 

7 

18 

Textile  drawings,  Reading  .  .  . 

91 

1 

Straight-faced  pulley . 

7 

14 

Threads . 

91 

30 

line,  A . 

6 

1 

Throw  of  a  cam . 

7 

36 

Subtraction . 

1 

12 

Thrown  silk . 

11 

15 

of  compound  quanti- 

Tight  and  loose  pulleys . 

7 

17 

ties . 

5 

13 

Time,  Measure  of . 

5 

7 

“  decimals . 

2 

22 

Toe  of  a  cam . 

7 

36 

11  fractions . 

2 

11 

Top  plan . 

91 

7 

Proof  of  . 

1 

14 

“  roll . 

15 

6 

Rule  for . 

1 

14 

Trains  of  gears . 

7 

29 

Sign  of . 

1 

12 

Transmission  of  power . 

8 

1 

Subtrahend . 

1 

12 

Power . 

7 

9 

Surds . 

4 

8 

Rope . 

7 

24 

Surface . 

6 

1 

“  “ 

8 

9 

Dark . 

91 

23 

Trapezium . 

6 

6 

“  Light . 

91 

23 

Trapezoid  . 

6 

5 

measure . 

5 

6 

Triangle . 

6 

2 

Surfaces,  Mensuration  of . 

6 

1 

Troy  weight . 

5 

3 

Surveyors’  measure . 

5 

5 

Twist  in  yarns . 

9 

18 

System  of  numbering  yarns,  Metric 

11 

38 

Twisted  yarns . 

9 

7 

Systems  of  numbering  single  yarns 

10 

13 

“  “ 

10 

7 

H  t< 

11 

20 

T 

Types  of  hangers . 

7 

‘  7 

Table  method  of  extracting  roots  . 

4 

16 

U 

“  Multiplication . 

1 

17 

of  angular  measure . 

5 

7 

Unit,  A . 

1 

1 

apothecaries’  fluid  meas- 

“  method  of  proportion  .... 

3 

18 

ure.  .  .  . 

5 

4 

“  of  a  number . 

1 

1 

weight .  .  . 

5 

4 

United  States  money . 

5 

2 

“  avoirdupois  weight .  .  .  . 

5 

3 

“  Reduction  of 

5 

11 

breaking  weight  of  warp 

Unlike  numbers . 

1 

1 

yarns . 

9 

20 

cubic  measure . 

5 

7 

V 

“  distance  between  hangers 

7 

7 

Value,  Local . 

1 

3 

“  “  drafts . 

15 

24 

“  of  a  fraction . 

2 

2 

“  dry  measure . 

5 

5 

“  “  “  ratio . 

3 

11 

“  “  linear  measure . 

5 

5 

“  Simple . 

1 

2 

INDEX 


XXV 


Sec. 


Vegetable  fibers . 11 

Velocity  of  rotating  pulleys  ....  8 

Vertical  line,  A .  6 

Views  and  their  arrangement  ...  91 

Diagrammatic . 91 

“  Perspective .  .  .  91 

“  Sectional . 91 

“  Side . 91 

Vinculum .  3 

Volume  of  a  cone  or  pyramid  ...  6 

“  cylinder .  6 

“  prism .  6 

“  “  sphere .  6 

“  the  frustum  of  a  pyra¬ 
mid  or  cone .  6 

“  or  solidity.  Measures  of  5 

W 

Wall-box  hanger .  7 

“  brackets .  7 

W  arp  . 11 

“  . 11 

. 13 

“  and  filling,  Counts  of  the  .  .  13 
“  Average  number  of  the  ...  10 
“  Finding  the  number  of  ends 

in  a . 12 

“  Number  of  ends  in  the  ...  13 

“  Pattern  of  the .  9 

“  “  "  “  . 10 

. . 11 

“  Weight  of  the . 13 

“  “  “  “  13 

“  yarn . 11 

“  “  . 11 

“  Breaking  weight  of  cot¬ 
ton  .  9 

“  Finding  hanks  of  ...  12 

“  Finding  number  of 

hanks  of . 12 

“  yarns,  Breaking  weight  of 

American .  9 

“  Calculations  for  .  .  12 

Warps,  Fancy .  9 

“  “  . 10 

“  “  . 11 

Weaving,  Contraction  during  ...  12 

“  “  in . 13 

Weight  and  counts.  Proof  of  ...  13 

“  Apothecaries’ .  5 

“  Avoirdupois  .  .  , .  5 

“  Measures  of .  5 

“  of  cloth . 13 

“  cotton  cloth . 12 

“  cloth  from  loom  ....  13 

“  “  filling . 13 


Sec.  Page 


Weight  of  the  filling . 

13 

23 

warp . 

13 

23 

warp . 

woolen  and  worsted 

13 

25 

cloth  . 

13 

3 

“  Troy  . 

5 

3 

Wheels,  Sprocket . 

Woolen  and  worsted  cloth  calcula- 

7 

31 

tions  .  .  . 

13 

1 

“  ply  yarns  .  . 

“  yarn  calcula- 

11 

26 

tions  .  .  . 

10 

1 

“  roving  and  yarn,  Sizing  .  . 

10 

15 

“  “  “  44  4  4 

11 

9 

yarns . 

“  Cut  system  of  num- 

11 

7 

bering . 

“  Cut  system  of  num- 

10 

14 

bering . 

“  Run  system  of  num- 

11 

9 

bering . 

“  Run  system  of  num- 

10 

13 

bering . 

11 

7 

Woolly  yarns . 

10 

13 

Worm,  A . 

7 

33 

“  gear  . 

7 

33 

“  gears.  Speed  of . 

Worsted  and  woolen  cloth  calcula- 

8 

12 

tions  .  .  . 

13 

1 

“  ply  yarns  .  . 
“  yarn  calcula- 

11 

26 

tions  .  .  . 

10 

1 

roving,  Sizing . 

10 

6 

“  “  “ 

11 

13 

yarns  . 

10 

1 

II  14 

11 

11 

“  Sizing . 

10 

3 

44  44  44 

11 

12 

Wrap  reel . 

9 

5 

4  4  4  4 

10 

4 

Y 

11 

5 

Yard  cloth . 

12 

2 

Yards  per  pound  . 

12 

21 

“  Finding . 

12 

13 

Yarn  and  roving.  Sizing . 

9 

3 

. .  “  cotton  .  . 

11 

4 

“  woolen  .  . 

10 

15 

44  44  44  44  44 

11 

9 

“  Breaking  weight  of  cotton  . 

9 

19 

“  calculations,  Cotton . 

9 

1 

for  filling .... 

12 

10 

General  .... 
“  “  Woolen  and 

11 

1 

worsted  .  .  . 

10 

1 

Page 

1 

5 

1 

3 

20 

18 

11 

6 

19 

14 

13 

11 

15 

14 

6 

9 

7 

30 

35 

2 

22 

23 

5 

22 

17 

22 

35 

23 

25 

30 

35 

19 

18 

7 

20 

5 

16 

21 

35 

8 

9 

22 

4 

3 

3 

21 

2 

24 

25 


XXVI 


INDEX 


Sec.  Page 

Sec.  Page 

Yarn,  Finding:  hanks  of  warp  .  .  . 

12 

18 

Yarns, 

Metric  system  of  number- 

“ 

number  of  hanks  of 

ing . 

10 

25 

warp . 

12 

7 

44 

Metric  system  of  number- 

“ 

numbering:  system . 

11 

2 

ing . 

11 

38 

“ 

Warp . 

11 

30 

i  4 

of  different  counts,  Folded  . 

9 

9 

“ 

4  1 

11 

35 

1 1 

44  44  44  44 

10 

9 

Yarns,  Average  number  of  .  .  .  . 

9 

15 

« ( 

44  44  44  44 

11 

22 

4  t 

44  44  “ 

11 

33 

« t 

materials.  Ply  . 

•11 

27 

“ 

Beamed . 

11 

30 

<  < 

“  spun  silk,  Ply . 

11 

26 

Breaking:  weight  of  ...  . 

9 

20 

« ( 

“  the  same  counts,  Folded 

9 

8 

“ 

Calculation  of  cost  of  ply  . 

11 

25 

<  ( 

4444  44  44  44 

10 

8 

44 

Calculations  for  warp  . 

12 

5 

t  < 

44  44  44  44  44 

11 

20 

“ 

of  cotton  ply  . 

11 

20 

4  4 

Ply . 

9 

7 

“ply . 

9 

8 

4  4 

44 

10 

7 

“ 

“  "  “ 

10 

8 

4  4 

‘  ‘ 

10 

17 

single  .  .  . 

9 

2 

4  4 

44 

11 

20 

44 

“  single  cot- 

4  4 

Run  system  of  numbering 

ton  .  .  . 

11 

3 

woolen . 

10 

13 

4  l 

“  single  wors- 

4  4 

Run  system  of  numbering 

ted  .  .  . 

10 

2 

woolen . 

11 

7 

composed  of  more  than  two 

4  4 

Silk . 

11 

14 

threads,  Ply .  . 

9 

9 

4  4 

Single . 

9 

1 

of  more  than  two 

4  4 

44 

11 

2 

threads,  Ply.  . 

10 

10 

4  4 

Sizing  worsted . 

10 

3 

41 

of  more  than  two 

4  4 

4  4  4  4 

11 

12 

threads,  Ply  .  . 

11 

23 

4  4 

System  of  numbering  single 

1  1 

Cotton . 

11 

2 

worsted . 

10 

1 

4i 

Cut  system  of  numbering 

4  4 

Systems  of  numbering  sin- 

woolen  . 

10 

14 

gle . 

10 

13 

41 

“  numbering 

4  4 

Twisted . 

9 

7 

woolen  . 

11 

9 

4  4 

4  4 

10 

7 

Diameter  of . 

11 

29 

4  4 

“ 

11 

20 

Folded . 

9 

7 

4  4 

Twist  in . 

9 

18 

“ 

(  i 

10 

7 

4  4 

Woolen . 

11 

7 

1  ‘ 

11 

20 

4  4 

“  and  worsted  ply  .  . 

11 

26 

Method  of  numbering  ply 

9 

7 

4  4 

Woolly . 

10 

13 

44  “  4  4  4  4 

10 

7 

4  4 

Worsted . 

10 

1 

n  il  a  il 

11 

20 

4  4 

4  4 

11 

11 

Methods  of  determining  the 

diameter  of . 

11 

29 

z 

4‘ 

Metric  systemof  numbering 

9 

21 

Zero 

1 

2 

